case study free fall

Free Fall Motion: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: February 16, 2023

Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex. We will explain many examples so you can see how to solve different types of projectile motion. 

What We Review

An object that is moving under only the influence of gravity is in free fall. In order for an object to be in free fall, wind and air resistance must be ignored. On Earth, all objects in free fall accelerate downward at the rate of gravity or 9.81\text{ m/s}^2 .

Applying Free Fall to Kinematic Equations

When analyzing free fall motion, we can apply the same kinematic equations as we did for motion on the ground. We can then use these equations to determine properties such as distance, time, and velocity. 

How to Find Distance Fallen for an Object in Free Fall

If an object is in free fall, we can use kinematic equations to find the distance it falls during a certain time. You will typically use the following kinematic equation to calculate the distance fallen:

In order to use this equation, you need to know the initial velocity of the object and the time of flight. Remember that the acceleration of a free falling object is always equal to the acceleration due to gravity, 9.81\text{ m/s}^2 . 

Many free fall physics problems will include scenarios where objects are dropped from rest. In this case, the initial velocity is zero and the first term of the kinematic equation above will cancel out. 

If the time is not known, another method for calculating the distance fallen is to use the following kinematic equation:

In this case, you must know the final velocity v_f of the object. Then, you can solve the equation for the distance d .

How to Find Time for an Object in Free Fall

The amount of time an object is in free fall will depend on its velocity and the distance it falls. Similar to distance, there are two equations you can use to find the time, depending on what you know. 

If you know the initial and final velocity of the object, then the simplest way to calculate time is using the kinematic equation:

This equation can be solved for time. Then, you’ll only need to substitute the values for the velocities and the acceleration due to gravity.

Another method to find time if you do not know the object’s final velocity is to use the equation:

Note that in this equation there are two terms that include the time t . Unless the initial velocity is zero, this can make it more challenging to solve this equation for time. If using this equation, you may need to use the quadratic formula to solve for time.

How to Find Final Velocity for an Object in Free Fall

The final velocity of an object in free fall depends on the amount of time it falls. Due to the acceleration of gravity, the velocity will increase every second by 9.81\text{ m/s} . The final velocity can be calculated using the equation:

If you do not know the amount of time the object is falling, another method for calculating the final velocity is using the kinematic equation: 

This equation requires that you instead know the distance that the object falls. If you are using this equation to find the final velocity, remember that the final velocity is squared in this equation. That means you will need to take a square root as your final step to solve for the final velocity. 

Examples of Free Fall

In this next section, we’ll apply the methods you just learned to solve some problems about free fall motion.

Example 1: How to Find the Distance for an Object Dropped from Rest

For example, an object is dropped from rest from the top of a tall building. It hits the ground 5\text{ s} after it is dropped. What is the height of the building? 

In this scenario, we know that the object’s initial velocity is zero because it was dropped from rest. We also know that the acceleration is 9.81\text{ m/s}^2 . This problem is asking us to find the distance the object falls. This will be equal to the height of the building.

Based on this information, we can use the following kinematic equation to find the distance:

Substituting the given values produces:

Therefore, the height of the building is about 123\text{ m} .

Example 2: How to Find the Final Velocity for an Object with Initial Velocity

In another example, an object in free fall has an initial, downward velocity of 2\text{ m/s} and falls a distance of 45\text{ m} . What is the object’s final velocity? 

In this scenario, we are given the object’s initial velocity, v_i and the distance d . We also know that the acceleration is 9.81\text{ m/s}^2 . Based on this information, we can use the following kinematic equation to find the final velocity:

Since the initial velocity is in the same direction as the acceleration (downward) we can use the same sign for both values.

Our last step is to eliminate the square by taking the square root:

Therefore, the final velocity of the object is about 30\text{ m/s} .

Motion Graphs for Objects in Free Fall

In addition to using physics equations, we can also represent free fall motion with motion graphs. Position-time graphs, velocity-time graphs, and acceleration-time graphs can tell us a lot about the object’s motion over time. Want a more in-depth review of motion graphs? Check out this blog post !

Position-Time Graph for an Object in Free Fall

In terms of position, many objects in free fall start at a high position, or height off the ground, and move downward. Objects in free fall accelerate due to gravity. Therefore, the position-time graph for free fall motion must be curved. This means that objects in free fall start with a slow velocity and gradually speed up which is represented by the steep downward curve of the graph. 

Velocity-Time Graph for an Object in Free Fall

As an object falls, its velocity increases due to the acceleration of gravity. This means that the velocity starts slow and steadily increases in the downward direction. The graph below shows the velocity-time for an object in free fall:

Note that the slope of this graph is constant and represents the acceleration due to gravity, or -9.81\text{ m/s}^2 .

Acceleration-Time Graph for an Object in Free Fall

Free fall acceleration is constant. Throughout the entire time that an object is falling, it is accelerating at a rate equal to the acceleration due to gravity, -9.81\text{ m/s}^2 . As shown in the graph below, the acceleration-time graph is a constant negative line. 

Projectile Motion

A projectile is an object that is launched or thrown into the air and then only influenced by gravity. Projectile motion has many similarities to free fall motion, however, projectiles may also travel a horizontal distance in addition to falling vertically down. 

Examples of Projectile Motion

The exact trajectory, or path, a projectile will take depends on how it is launched. However, all projectiles follow a curved trajectory such as in the image shown below:

If you play or watch sports, you likely have already observed projectile motion. Projectile motion describes the arc of a basketball in a free throw, a fly ball in baseball, or a volleyball bumped over the net. 

Horizontal Component of Velocity

To analyze projectile motion, we must separate the motion into horizontal and vertical components. The horizontal component of a projectile’s velocity is independent of the vertical component of velocity. Since gravity acts vertically, there are no horizontal forces acting on projectiles. This means that the horizontal component of a projectile’s velocity remains constant throughout the entire flight. 

Example: Finding the Horizontal Component

For example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the horizontal component of the projectile’s velocity?

We will need to use trig identities to determine the components of the velocity. We can visualize the components as a triangle where the hypotenuse is the initial velocity and the sides represent the horizontal, v_{ix} , and vertical, v_{iy} , components of the velocity.

Cosine is defined as the adjacent side of the triangle divided by the hypotenuse. Since the horizontal component is adjacent to the angle, we can use cosine to find the horizontal component of velocity:

Therefore, the horizontal component of the initial velocity is 4\text{ m/s} .

Need to review your trig identities? Try out this resource from Khan Academy .

Vertical Component of Velocity

The vertical component of a projectile’s velocity will be influenced by gravity, which acts vertically on the object causing it to accelerate downward. Therefore, the vertical component of velocity will change throughout the projectile’s flight. We can calculate the vertical component of velocity at a particular time in a method similar to calculating the horizontal component. 

Example: Finding the Vertical Component

In the same example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the vertical component of the projectile’s velocity?

As we visualize the velocity components, we are solving this time for the opposite side of the triangle. Sine is defined as the opposite side of the triangle divided by the hypotenuse. Therefore, the initial vertical velocity is:

Solving Projectile Motion Questions

Let’s apply what we’ve learned to some examples of projectile motion!

Example 1: Finding the Range of a Projectile

In this example, a projectile is fired horizontally with a speed of 5\text{ m/s} from a cliff with a height of 60\text{ m} . How far from the base of the cliff will the projectile land? 

In this scenario, we are given the initial horizontal velocity v_{ix}=5\text{ m/s} and the vertical change in position d_y=-60\text{ m} . Since the projectile is launched horizontally, the initial vertical velocity, v_{iy} , is zero. We also always know in projectile motion that the vertical acceleration is a_y=-9.81\text{ m/s}^2 and the horizontal acceleration, a_x , is zero.

This problem is asking us to find the horizontal displacement, or d_x . This is also referred to as the range . We can use the following kinematic equation to find the projectile’s final horizontal position:

Since the horizontal acceleration of a projectile is zero, this equation can be simplified to:

Before we can solve this equation, we must first determine the time of the projectile’s flight. We can actually use this same equation in the vertical direction to solve for time:

Since the initial vertical velocity is zero, this equation can be simplified to:

Solving for t :

Substituting the given values:

Now we can use this time to calculate the horizontal displacement of the projectile:

Therefore, the projectile will land about 17.5\text{ m} from the base of the cliff. 

Example 2: Finding the Maximum Height of a Projectile

As another example, a projectile is launched from the ground with an initial velocity of 25\text{ m/s} at an angle of 50^{\circ} . What is the projectile’s maximum height?

As a projectile travels upward, its vertical velocity becomes slower and slower due to the negative acceleration of gravity. At the maximum height of the trajectory, the projectile’s vertical velocity will momentarily be zero as the projectile stops and turns to move downward. Therefore, in this scenario, our final vertical velocity, v_{fy} , is zero.

We can use the following kinematic equation to solve for the maximum height, d_y :

Solving for d_y :

Before we can use this equation to calculate the height, we will need to use the sine trig identity to find the vertical component of the initial velocity:

Since the initial velocity is in the opposite direction as the acceleration, it’s really important to remember the sign here. If we define moving up as positive, then the initial velocity is positive and the acceleration is negative. Substituting this initial vertical velocity and the given values into the equation above gives:

Therefore, the projectile will reach a maximum height of about 18.7\text{ m} .

For more examples and an explanation of solving these types of projectile motion problems, check out this youtube video from Professor Dave . 

Understanding free fall and projectile motion allows you to solve some of the most complex problems you will encounter in introductory physics. All projectiles are acted on only by gravity, and the vertical and horizontal components of motion are independent of each other. This allows us to apply our kinematic equations to solve for a projectile’s time of flight, velocity, and displacement in each direction.  

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• Earth and space

Free Fall – Introduction, Explanation & Case Study

Key concepts:.

• Mass and Weight
• Weightlessness

Introduction: 

The mass of an object tells us how much matter it is composed of. It is measured in kilogram. The weight of an object is the force of gravity that acts on it and is measured in Newton.  

The formula of weight: 

Force = Mass × Acceleration 

Weight = Mass × Gravitational acceleration  

Gravitational Acceleration = g  

(Acceleration caused by the pull of gravitational force) 

Gravitational Acceleration = g = GM / R 2

Here G = Gravitational Constant 

M = Mass of the planet 

R = Radius of the Planet 

There are two types of forces contact forces and non-contact forces. The contact forces are frictional force , tension force, and normal reaction or support force. The non-contact forces are Gravitational force, electric force, and magnetic force. 

Explanation: 

We can feel contact forces but not gravity because gravity is a non-contact force, 

Freefall is a kind of sensation in which the object doesn’t feel the effect of gravity in absence of contact forces. When the only force acting on an object is gravity, we say that the object is falling freely. 

Weight is the force with which the Earth attracts us. We are conscious of our own weight when we are standing on a surface since the surface applies a force opposite to our weight to keep us at rest. 

A man standing on floor is in equilibrium under the action of two forces : weight w and normal reaction N. 

ΣF = N – W = 0 

The same principle holds good when we measure the weight of an object by a spring balance hung from a fixed point like a ceiling. When we hang an object to a spring balance, the spring is pulled down a little by the gravitational pull of the object and in turn, the spring exerts a force on the object vertically upwards. 

Now imagine the top end of the spring balance is no longer fixed to the ceiling both ends of the spring as well as the object move with the same acceleration g. The spring is not stretched and does not exert any upward force on the object which is moving down with acceleration g due to gravity. The reading recorded in the spring balance will be zero.  

If the object were a human being, he or she will not feel his weight since there is no upward force on him. Thus, when an object is in free fall it is weightless. 

In an elevator, the forces acting on an object with a mass m are: 

The gravitational force (mg) in the downward direction 

The reaction force N or push of the elevator floor on the object (N) in the upward direction. 

Acceleration of Lift = a.  

The weighing scale reading: 

When the lift is at rest or in uniform motion  

Acceleration = a = 0  

Net force on lift = F = N – mg = 0 

But N = reading of the weighting scale = W 

When the lift is moving downwards with acceleration = a 

Then mg – N will provide downward acceleration   

Net force on lift = F = mg – N = ma 

N = mg – ma  

W = mg – ma 

When the lift is moving upwards with acceleration = a 

Then N – mg will provide upward acceleration   

Net force on lift = F = N – mg = ma 

N = mg + ma 

W = mg + ma 

When the lift is under freefall 

moving downwards with acceleration = g 

Then N – mg will provide downward acceleration g   

Net force on lift = F = N – mg = mg 

But N = rearing of the weighting scale = W 

Weightlessness: A freely falling object experiences weightlessness as the only force acting on it is the force of gravity which acts downwards. The object’s acceleration is equal to the acceleration due to gravity. Hence, the net force acting on the object is zero. 

The apparent weight we feel while riding a roller coaster along a curved path arises from the support we get from the floor and seat. When we ride a roller coaster towards the top, the downward acceleration of our seats is equal to the acceleration due to gravity. As a result, we experience weightlessness since there is no net force acting on us during free-fall. 

Another situation when we feel weightless is during the take-off and landing of an airplane. When the airplane accelerates or decelerates at a rate equal to the acceleration due to gravity, we experience weightlessness. 

Weightlessness time duration:  

During High Jump or Long Jump: 1 Sec 

Jumping from a Tower: 1-2 Sec 

Freefall ride: 2-3 sec 

During Airplane landing or take-off (Or Sky Diving): 10-30sec 

• The difference between  Weight and Mass  s that the mass of an object tells us how much matter it is composed of whereas the weight of an object is the force of gravity that acts on it. Mass is measured in kilogram and weight is measured in Newton.
• We can feel contact forces but not gravity because gravity is a non-contact force.
• Free fall is a kind of sensation in which the object doesn’t feel the effect of gravity in absence of contact forces.
• In an elevator, the forces acting on an object with a mass m are the gravitational force(mg) in the downward direction and the reaction force or push of the elevator floor on the object (N = ma) in the upward direction. Acceleration of Lift = a.
• Weighing Scale Reading:

Case 1:   When the lift is at rest or in uniform motion Acceleration = a =0, W = mg,

Case 2:  When the lift is moving down with acceleration a. (a – acting downward) W = mg – ma Lightweight

Case 3:  When the lf is moving upward with acceleration a. (a – acting upward) W = mg + ma Heavy Weight

Case 4:  When the lift is falling freely a= g (Free Fall) W = 0 Weightlessness

• Weightlessness:  A freely falling object experiences weightlessness as the only force acting on it is the force of gravity which acts downwards. The object’s acceleration is equal to the acceleration due to gravity. Hence, the net force acting on the object s zero.
• A freely falling object experiences weightlessness as the only force acting on it is the force of gravity which acts downwards. The object’s acceleration is equal to the acceleration due to gravity. Hence, the net force acting on the object is zero.

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3.6: Free Fall

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Learning Objectives

• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.
• Use one-dimensional motion in perpendicular directions to analyze projectile motion.
• Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
• Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
• Calculate the trajectory of a projectile.

An interesting application of Equation 3.3.2 through Equation 3.5.22 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure \(\PageIndex{1}\).

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location on Earth and has the average value

\[g = 9.81\; m/s^{2}\; (or\; 32.2\; ft/s^{2}) \ldotp\]

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value +g or -g depends on how we define our coordinate system. If we define the upward direction as positive, then a = -g = -9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 .

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g. We represent vertical displacement with the symbol y.

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals -g (with the positive direction upward).

\[v =v _{0} - gt \label{3.15}\]

\[y = y_{0} + v_{0} t - \frac{1}{2} gt^{2} \label{3.16}\]

\[v^{2} = v_{0}^{2} - 2 g(y - y_{0}) \label{3.17}\]

Problem-Solving Strategy: Free Fall

• Decide on the sign of the acceleration of gravity. In Equation \ref{3.15} through Equation \ref{3.17}, acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
• Draw a sketch of the problem. This helps visualize the physics involved.
• Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
• Decide which of Equation \ref{3.15} through Equation \ref{3.17} are to be used to solve for the unknowns.

Example \(\PageIndex{1}\): Free Fall of a Ball

Figure \(\PageIndex{2}\) shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that is thrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b) What is the velocity when it arrives at the ground?

Choose the origin at the top of the building with the positive direction upward and the negative direction downward. To find the time when the position is -98 m, we use Equation \ref{3.16}, with y 0 = 0, v 0 = -4.9 m/s, and g = 9.8 m/s 2 .

• Substitute the given values into the equation: \(y = y_{0} + v_{0} t - \frac{1}{2} gt^{2}\) \(-98.0\; m = 0 - (4.9\; m/s)t - \frac{1}{2} (9.8\; m/s^{2}) t^{2} \ldotp\) This simplifies to \(t^{2} + t - 20 = 0 \ldotp\) This is a quadratic equation with roots t = -5.0 s and t = 4.0 s. The positive root is the one we are interested in, since time t = 0 is the time when the ball is released at the top of the building. (The time t = -5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
• Using Equation \ref{3.15}, we have \(v =v _{0} - gt = -4.9\; m/s - (9.8\; m/s^{2})(4.0\; s) = -44.1\; m/s \ldotp\)

Significance

For situations when two roots are obtained from a quadratic equation in the time variable, we must look at the physical significance of both roots to determine which is correct. Since t = 0 corresponds to the time when the ball was released, the negative root would correspond to a time before the ball was released, which is not physically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ball interacts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zero velocity. This problem shows how important it is to establish the correct coordinate system and to keep the signs of g in the kinematic equations consistent.

Example \(\PageIndex{2}\): Vertical Motion of a Baseball

A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck Figure \(\PageIndex{3}\). (a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does it take to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity of the ball when it is caught? Assume the ball is hit and caught at the same location.

Choose a coordinate system with a positive y-axis that is straight up and with an origin that is at the spot where the ball is hit and caught.

• Equation \ref{3.16} gives \(y = y_{0} + v_{0} t - \frac{1}{2} gt^{2}\) \(0 = 0 + v_{0} (5.0\; s)- \frac{1}{2} (9.8\; m/s^{2}) (5.0\; s)^{2} \ldotp\) which gives v 0 = 24.5 m/sec.
• At the maximum height, v = 0. With v 0 = 24.5 m/s, Equation \ref{3.17} gives \(v^{2} = v_{0}^{2} - 2 g(y - y_{0})\) \(0 = (24.5\; m/s^{2}) - 2 (9.8\; m/s^{2})(y - 0)\) or \(y = 30.6\; m \ldotp\)
• To find the time when v = 0 , we use Equation \ref{3.15}: \(v = v_{0} - gt\) \(0 = 24..5\; m/s - (9.8\; m/s^{2})t \ldotp\) This gives t = 2.5 s. Since the ball rises for 2.5 s, the time to fall is 2.5 s.
• The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
• The velocity at t = 5.0 s can be determined with Equation \ref{3.15}: \(\begin{split} v & = v_{0} - gt \\ & = 24.5\; m/s - 9.8\; m/s^{2} (5.0\; s) \\ & = -24.5\; m/s \ldotp \end{split}\)

The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. We used a single equation to go from throw to catch, and did not have to break the motion into two segments, upward and downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It is important to understand, as illustrated in this example, that objects moving upward away from Earth are also in a state of free fall.

Example \(\PageIndex{3}\): Rocket Booster

A small rocket with a booster blasts off and heads straight upward. When at a height of 5.0 km and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.

We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. We are given the initial velocity of the booster and its height. We consider the point of release as the origin. We know the velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster is zero at its maximum height, so we can use this information as well. From these observations, we use Equation \ref{3.17}, which gives us the maximum height of the booster. We also use Equation \ref{3.17} to give the velocity at 6.0 km. The initial velocity of the booster is 200.0 m/s.

• From Equation \ref{3.17}, \(v^{2} = v_{0}^{2} - 2 g(y - y_{0})\). With v = 0 and y 0 = 0, we can solve for y: \(y = \frac{v_{0}^{2}}{-2g} = \frac{(2.0 \times 10^{2}\; m/s)^{2}}{-2(9.8\; m/s^{2})} = 2040.8\; m \ldotp\) This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
• An altitude of 6.0 km corresponds to y = 1.0 x 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0, and v 0 = 200.0 m/s. We have, from Equation \ref{3.17}, \(v^{2} = (200.0\; m/s)^{2} - 2(9.8\; m/s^{2})(1.0 \times 10^{3}\; m) \Rightarrow v = \pm 142.8\; m/s \ldotp\)

We have both a positive and negative solution in (b). Since our coordinate system has the positive direction upward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of the trajectory of the booster. The value v = -142.8 m/s corresponds to the velocity at 6000 m on the downward leg. This example is also important in that an object is given an initial velocity at the origin of our coordinate system, but the origin is at an altitude above the surface of Earth, which must be taken into account when forming the solution.

Two-Dimensional Motion Involving Gravity-Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure \(\PageIndex{1}\) illustrates the notation for displacement, where we define \(\vec{s}\) to be the total displacement, and \(\vec{x}\) and \(\vec{y}\) are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

\[a_{y} = -g = -9.8\; m/s^{2} (- 32\; ft/s^{2}) \ldotp\]

Because gravity is vertical, a x = 0. If a x = 0, this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration. The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = -g, a x = 0:

Horizontal Motion

\[v_{0x} = v_{x}, \quad x = x_{0} + v_{x} t \label{4.19}\]

Vertical Motion

\[y = y_{0} + \frac{1}{2} (v_{0y} + v_{y})t \label{4.20}\]

\[v_{y} = v_{0y} - gt \label{4.21}\]

\[y = y_{0} + v_{0y} t - \frac{1}{2} g t^{2} \label{4.22}\]

\[v_{y}^{2}= v_{0y}^{2} + 2g(y - y_{0}) \label{4.23}\]

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

• Resolve the motion into horizontal and vertical components along the x- and y-axes. The magnitudes of the components of displacement \(\vec{s}\) along these axes are x and y. The magnitudes of the components of velocity \(\vec{v}\) are v x = vcos\(\theta\) and v y = vsin\(\theta\), where v is the magnitude of the velocity and \(\theta\) is its direction relative to the horizontal, as shown in Figure \(\PageIndex{2}\).
• Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
• Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
• Recombine quantities in the horizontal and vertical directions to find the total displacement \(\vec{s}\) and velocity \(\vec{v}\). Solve for the magnitude and direction of the displacement and velocity using \(s = \sqrt{x^{2} + y^{2}} \ldotp \quad \phi = \tan^{-1} \left(\dfrac{y}{x}\right), \quad v = \sqrt{v_{x}^{2} + v_{y}^{2}} \ldotp\) where \(\phi\) is the direction of the displacement \(\vec{s}\).

Example \(\PageIndex{4}\): A Fireworks Projectile Explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure \(\PageIndex{3}\). The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point?

The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = -g. We can then define x 0 and y 0 to be zero and solve for the desired quantities.

• By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0. Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y: \(v_{y}^{2} = v_{0y}^{2} - 2g(y - y_{0}) \ldotp\) Because y 0 and v y are both zero, the equation simplifies to \(0 = v_{0y}^{2} - 2gy \ldotp\) Solving for y gives \(y = \frac{v_{0y}^{2}}{2g} \ldotp\) Now we must find v 0y , the component of the initial velocity in the y direction. It is given by v 0y = v 0 sin\(\theta_{0}\), where v 0 is the initial velocity of 70.0 m/s and \(\theta_{0}\) = 75° is the initial angle. Thus \(v_{0y} = v_{0} \sin \theta = (70.0\; m/s) \sin 75^{o} = 67.6\; m/s\) and y is \(y = \frac{(67.6\; m/s)^{2}}{2(9.80\; m/s^{2})} \ldotp\) Thus, we have \(y = 233\; m \ldotp\) Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.
• As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0y - gt. Because v y = 0 at the apex, this equation reduces \(0 = v_{0y} - gt\) or \(t = \frac{v_{0y}}{g} = \frac{67.6\; m/s}{9.80\; m/s^{2}} = 6.90\; s \ldotp\) This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + \(\frac{1}{2}\)(v 0y + v y )t. This is left for you as an exercise to complete.
• Because air resistance is negligible, a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t, where x 0 is equal to zero. Thus, \(x = v_{x} t,\) where v x is the x-component of the velocity, which is given by \(v_{x} = v_{0} \cos \theta = (70.0\; m/s) \cos 75^{o} = 18.1\; m/s \ldotp\) Time t for both motions is the same, so x is \(x = (18.1\; m/s)(6.90\; s) = 125\; m \ldotp\) Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.
• The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point: \(\vec{s} = 125 \hat{i} + 233 \hat{j}\) \(|\vec{s}| = \sqrt{125^{2} + 233^{2}} = 264\; m\) \(\theta = \tan^{-1} \left(\dfrac{233}{125}\right) = 61.8^{o} \ldotp\) Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure \(\PageIndex{1}\), which shows the curvature of the trajectory toward the ground level. When solving Example 4.7(a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h. Then, \(h = \frac{v_{0y}^{2}}{2g} \ldotp\) This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Example \(\PageIndex{5}\): Calculating projectile motion- Tennis Player

A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal (Figure \(\PageIndex{4}\)). On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball’s velocity at impact?

Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain \(\vec{v}\) at final time t, determined in the first part of the example.

• While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation \ref{4.22}: \(y = y_{0} + v_{0y}t - \frac{1}{2} gt^{2} \ldotp\) If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: \(v_{0y} = v_{0} \sin \theta_{0} = (30.0\; m/s) \sin 45^{o} = 21.2\; m/s \ldotp\) Substituting into Equation \ref{4.22} for y gives us \(10.0\; m = (21.2\; m/s)t - (4.90\; m/s^{2})t^{2} \ldotp\) Rearranging terms gives a quadratic equation in t: \((4.90\; m/s^{2})t^{2} - (21.2\; m/s)t + 10.0\; m = 0 \ldotp\) Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator: \(t = 3.79\; s \ldotp\) The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
• We can find the final horizontal and vertical velocities v x and v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector \(\vec{v}\) and the angle \(\theta\) it makes with the horizontal. Since v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore, \(v_{x} = v_{0} \cos \theta_{0} = (30\; m/s) \cos 45^{o} = 21.2\; m/s \ldotp\) The final vertical velocity is given by Equation \ref{4.21}: \(v_{y} = v_{0y} - gt \ldotp\) Since \(v_{0y}\) was found in part (a) to be 21.2 m/s, we have \(v_{y} = 21.2\; m/s - (9.8\; m/s^{2})(3.79 s) = -15.9\; m/s \ldotp\) The magnitude of the final velocity \(\vec{v}\) is \(v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(21.2\; m/s)^{2} + (-15.9\; m/s)^{2}} = 26.5\; m/s \ldotp\) The direction \(\theta_{v}\) is found using the inverse tangent: \(\theta_{v} = \tan^{-1} \left(\dfrac{v_{y}}{v_{x}}\right) = \tan^{-1} \left(\dfrac{21.2}{-15.9}\right) = -53.1^{o} \ldotp\)
• As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
• The negative angle means the velocity is 53.1° below the horizontal at the point of impact. This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation.

Example \(\PageIndex{5}\): Time of Flight for a Projectile on a Horizontal Surface

Solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface.

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

\[y - y_{0} = v_{0y} t - \frac{1}{2} gt^{2} = (v_{0} \sin \theta_{0})t - \frac{1}{2} gt^{2} = 0 \ldotp\]

Factoring, we have

\[t \left(v_{0} \sin \theta_{0} - \dfrac{gt}{2}\right) = 0 \ldotp\]

Solving for t gives us

\[t_{tof} = \frac{2(v_{0} \sin \theta_{0})}{g} \ldotp \label{4.24}\]

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation \ref{4.24} does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

Example \(\PageIndex{5}\): Trajectory of a Projectile on a Horizontal Surface

Find an expression for the trajectory of a projectile

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(x). We take x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

\[x = v_{0x}t \Rightarrow t = \frac{x}{v_{0x}} = \frac{x}{v_{0} \cos \theta_{0}} \ldotp\]

Substituting the expression for t into the equation for the position y = (v 0 sin \(\theta_{0}\))t - \(\frac{1}{2}\) gt 2 gives

\[y = (v_{0} \sin \theta_{0}) \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right) - \frac{1}{2} g \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right)^{2} \ldotp\]

Rearranging terms, we have

\[y = (\tan \theta_{0})x - \Big[ \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \Big] x^{2} \ldotp \label{4.25}\]

This trajectory equation is of the form y = ax + bx 2 , which is an equation of a parabola with coefficients

\[a = \tan \theta_{0}, \quad b = - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \ldotp\]

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure \(\PageIndex{7}\), which is based on a drawing in Newton’s Principia . If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in another chapter.

3.11: Free-falling Bodies: Introduction

Chapter 1: units, dimensions, and measurements, chapter 2: vectors and scalars, chapter 3: motion along a straight line, chapter 4: motion in two or three dimensions, chapter 5: newton's laws of motion, chapter 6: application of newton's laws of motion, chapter 7: work and kinetic energy, chapter 8: potential energy and energy conservation, chapter 9: linear momentum, impulse and collisions, chapter 10: rotation and rigid bodies, chapter 11: dynamics of rotational motions, chapter 12: equilibrium and elasticity, chapter 13: fluid mechanics, chapter 14: gravitation, chapter 15: oscillations, chapter 16: waves, chapter 17: sound, chapter 18: temperature and heat, chapter 19: the kinetic theory of gases, chapter 20: the first law of thermodynamics, chapter 21: the second law of thermodynamics, chapter 22: electric charges and fields, chapter 23: gauss's law, chapter 24: electric potential, chapter 25: capacitance, chapter 26: current and resistance, chapter 27: direct-current circuits, chapter 28: magnetic forces and fields, chapter 29: sources of magnetic fields, chapter 30: electromagnetic induction, chapter 31: inductance, chapter 32: alternating-current circuits, chapter 33: electromagnetic waves.

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When an object falls towards the ground and is only influenced by gravity, it is said to be in free-fall.

Consider a ball thrown upwards in the air, reaches a certain height before falling back due to Earth's gravitational force.

Here, the ball is undergoing changes in velocity during its fall, but the acceleration is constant.

Consider a paper and a stone both being dropped at the same time and from the same height. Which object will reach the ground first?

It is observed that the stone reaches the ground first, followed by paper. This is due to the air resistance, which is negligible for the heavy objects.

However, if the objects are dropped together from the same height in a closed vacuum box, they will reach the base of the box at the same time.

Therefore, if the air effects are ignored, all objects irrespective of their masses, fall with a constant acceleration, known as acceleration due to gravity, which has an approximate value of 9.8 m/s 2 .

All objects, neglecting air resistance, fall with the same acceleration towards the Earth's center due to the force exerted by the Earth's gravity. This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavier ones. People believed that a heavier object had a greater acceleration when falling until Galileo Galilei (1564–1642) proved otherwise. We now know this is not the case.

The motion of an object under the influence of gravitational force is called a free-fall. Free-fall motion is not only associated with objects dropped from rest, but with all objects moving freely under gravity's influence. The acceleration of objects under a free-fall is constant, and is known as acceleration due to gravity, g . The magnitude of acceleration due to gravity is 9.8 m/s 2 . Because of the constant acceleration, kinematic equations of motion can be used to predict the dynamics of objects under free-fall, provided that the air effects are neglected.

In the real world, however, air resistance is always present, and it opposes the motion of objects. The momentum required to oppose the motion of heavier objects is larger compared to lighter objects. Therefore, heavier objects fall faster towards the ground during free-fall in the presence of air resistance.

This text is adapted from Openstax, University Physics Volume 1, Section 3.5: Free-fall .

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Describing Free Fall Video Tutorial

• What is meant by free fall or free-falling motion?
• How can the position, velocity, and acceleration be described during the course of a free fall motion?
• How can the velocity of an object be calculated at any given instant in time?

View Video Tutorial

Help us help you.

acceleration due to gravity

Want to see an object accelerate?

• Pick something up with your hand and drop it. When you release it from your hand, its speed is zero. On the way down its speed increases. The longer it falls the faster it travels. Sounds like acceleration to me.
• But acceleration is more than just increasing speed. Pick up this same object and toss it vertically into the air. On the way up its speed will decrease until it stops and reverses direction. Decreasing speed is also considered acceleration.
• But acceleration is more than just changing speed. Pick up your battered object and launch it one last time. This time throw it horizontally and notice how its horizontal velocity gradually becomes more and more vertical. Since acceleration is the rate of change of velocity with time and velocity is a vector quantity, this change in direction is also considered acceleration.

In each of these examples the acceleration was the result of gravity. Your object was accelerating because gravity was pulling it down. Even the object tossed straight up is falling — and it begins falling the minute it leaves your hand. If it wasn't, it would have continued moving away from you in a straight line. This is the acceleration due to gravity .

What are the factors that affect this acceleration due to gravity? If you were to ask this of a typical person, they would most likely say "weight" by which they actually mean "mass" (more on this later). That is, heavy objects fall fast and light objects fall slow. Although this may seem true on first inspection, it doesn't answer my original question. "What are the factors that affect the acceleration due to gravity ?" Mass does not affect the acceleration due to gravity in any measurable way. The two quantities are independent of one another. Light objects accelerate more slowly than heavy objects only when forces other than gravity are also at work. When this happens, an object may be falling, but it is not in free fall. Free fall occurs whenever an object is acted upon by gravity alone.

Try this experiment.

• Obtain a piece of paper and a pencil. Hold them at the same height above a level surface and drop them simultaneously. The acceleration of the pencil is noticeably greater than the acceleration of the piece of paper, which flutters and drifts about on its way down.

Something else is getting in the way here — and that thing is air resistance (also known as aerodynamic drag). If we could somehow reduce this drag we'd have a real experiment. No problem.

• Repeat the experiment, but before you begin, wad the piece of paper up into the tightest ball possible. Now when the paper and pencil are released, it should be obvious that their accelerations are identical (or at least more similar than before).

We're getting closer to the essence of this problem. If only somehow we could eliminate air resistance altogether. The only way to do that is to drop the objects in a vacuum. It is possible to do this in the classroom with a vacuum pump and a sealed column of air. Under such conditions, a coin and a feather can be shown to accelerate at the same rate. (In the olden days in Great Britain, a coin called a guinea was used and so this demonstration is sometimes called the "guinea and feather".) A more dramatic demonstration was done on the surface of the moon — which is as close to a true vacuum as humans are likely to experience any time soon. Astronaut David Scott released a rock hammer and a falcon feather at the same time during the Apollo 15 lunar mission in 1971. In accordance with the theory I am about to present, the two objects landed on the lunar surface simultaneously (or nearly so). Only an object in free fall will experience a pure acceleration due to gravity.

the leaning tower of Pisa

Let's jump back in time for a bit. In the Western world prior to the 16th century, it was generally assumed that the acceleration of a falling body would be proportional to its mass — that is, a 10 kg object was expected to accelerate ten times faster than a 1 kg object. The ancient Greek philosopher Aristotle of Stagira (384–322 BCE), included this rule in what was perhaps the first book on mechanics. It was an immensely popular work among academicians and over the centuries it had acquired a certain devotion verging on the religious. It wasn't until the Italian scientist Galileo Galilei (1564–1642) came along that anyone put Aristotle's theories to the test. Unlike everyone else up to that point, Galileo actually tried to verify his own theories through experimentation and careful observation. He then combined the results of these experiments with mathematical analysis in a method that was totally new at the time, but is now generally recognized as the way science gets done. For the invention of this method, Galileo is generally regarded as the world's first scientist.

In a tale that may be apocryphal, Galileo (or an assistant, more likely) dropped two objects of unequal mass from the Leaning Tower of Pisa. Quite contrary to the teachings of Aristotle, the two objects struck the ground simultaneously (or very nearly so). Given the speed at which such a fall would occur, it is doubtful that Galileo could have extracted much information from this experiment. Most of his observations of falling bodies were really of round objects rolling down ramps. This slowed things down enough to the point where he was able to measure the time intervals with water clocks and his own pulse (stopwatches and photogates having not yet been invented). This he repeated "a full hundred times" until he had achieved "an accuracy such that the deviation between two observations never exceeded one-tenth of a pulse beat."

With results like that, you'd think the universities of Europe would have conferred upon Galileo their highest honor, but such was not the case. Professors at the time were appalled by Galileo's comparatively vulgar methods even going so far as to refuse to acknowledge that which anyone could see with their own eyes. In a move that any thinking person would now find ridiculous, Galileo's method of controlled observation was considered inferior to pure reason. Imagine that! I could say the sky was green and as long as I presented a better argument than anyone else, it would be accepted as fact contrary to the observation of nearly every sighted person on the planet.

Galileo called his method "new" and wrote a book called Discourses on Two New Sciences wherein he used the combination of experimental observation and mathematical reasoning to explain such things as one dimensional motion with constant acceleration, the acceleration due to gravity, the behavior of projectiles, the speed of light, the nature of infinity, the physics of music, and the strength of materials. His conclusions on the acceleration due to gravity were that…

the variation of speed in air between balls of gold, lead, copper, porphyry, and other heavy materials is so slight that in a fall of 100 cubits a ball of gold would surely not outstrip one of copper by as much as four fingers. Having observed this I came to the conclusion that in a medium totally devoid of resistance all bodies would fall with the same speed. For I think no one believes that swimming or flying can be accomplished in a manner simpler or easier than that instinctively employed by fishes and birds. When, therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everybody? I greatly doubt that Aristotle ever tested by experiment. Galileo Galilei, 1638

Despite that last quote, Galileo was not immune to using reason as a means to validate his hypothesis. In essence, his argument ran as follows. Imagine two rocks, one large and one small. Since they are of unequal mass they will accelerate at different rates — the large rock will accelerate faster than the small rock. Now place the small rock on top of the large rock. What will happen? According to Aristotle, the large rock will rush away from the small rock. What if we reverse the order and place the small rock below the large rock? It seems we should reason that two objects together should have a lower acceleration . The small rock would get in the way and slow the large rock down. But two objects together are heavier than either by itself and so we should also reason that they will have a greater acceleration . This is a contradiction.

Here's another thought problem. Take two objects of equal mass. According to Aristotle, they should accelerate at the same rate. Now tie them together with a light piece of string. Together, they should have twice their original acceleration. But how do they know to do this? How do inanimate objects know that they are connected? Let's extend the problem. Isn't every heavy object merely an assembly of lighter parts stuck together? How can a collection of light parts, each moving with a small acceleration, suddenly accelerate rapidly once joined? We've argued Aristotle into a corner. The acceleration due to gravity is independent of mass.

Galileo made plenty of measurements related to the acceleration due to gravity but never once calculated its value (or if he did, I have never seen it reported anywhere). Instead he stated his findings as a set of proportions and geometric relationships — lots of them. His description of constant speed required one definition, four axioms, and six theorems. All of these relationships can now be written as the single equation in modern notation.

Algebraic symbols can contain as much information as several sentences of text, which is why they are used. Contrary to the common wisdom, mathematics makes life easier.

location, location, location

The generally accepted value for the acceleration due to gravity on and near the surface of the Earth is…

g  =  9.8 m/s 2

or in non-SI units…

g  =  35 kph/s = 22 mph/s = 32 feet/s 2

It is useful to memorize this number (as millions of people around the globe already have), however, it should also be pointed out that this number is not a constant . Although mass has no effect on the acceleration due to gravity, there are three factors that do. They are location, location, location.

Everyone reading this should be familiar with the images of the astronauts hopping about on the moon and should know that the gravity there is weaker than it is on the Earth — about one sixth as strong or 1.6 m/s 2 . That's why the astronauts were able to hop around on the surface easily despite the weight of their space suits. In contrast, gravity on Jupiter is stronger than it is on Earth — about two and a half times stronger or 25 m/s 2 . Astronauts cruising through the top of Jupiter's thick atmosphere would find themselves struggling to stand up inside their space ship.

On the Earth, gravity varies with latitude and altitude (to be discussed in a later chapter ). The acceleration due to gravity is greater at the poles than at the equator and greater at sea level than atop Mount Everest. There are also local variations that depend upon geology. The value of 9.8 m/s 2 — with only two significant digits — is true for all places on the surface of the Earth and holds for altitudes up to +10 km (the altitude of commercial jet airplanes) and depths down to −20 km (far below the deepest mines).

How crazy are you for accuracy? For most applications, the value of 9.8 m/s 2 is more than sufficient. If you're in a hurry, or don't have access to a calculator, or just don't need to be that accurate; rounding g on Earth to 10 m/s 2 is often acceptable. During a multiple choice exam where calculators aren't allowed, this is often the way to go. If you need greater accuracy, consult a comprehensive reference work to find the accepted value for your latitude and altitude.

If that's not good enough, then obtain the required instruments and measure the local value to as many significant digits as you can. You may learn something interesting about your location. I once met a geologist whose job it was to measure g across a portion of West Africa. When I asked him who he worked for and why he was doing this, he basically refused to answer other than to say that one could infer the interior structure of the Earth from a gravimetric map prepared from his findings. Knowing this, one might then be able to identify structures where valuable minerals or petroleum might be found.

Like all professions, those in the gravity measuring business ( gravimetry ) have their own special jargon. The SI unit of acceleration is the meter per second squared [m/s 2 ]. Split that into a hundred parts and you get the centimeter per second squared [cm/s 2 ] also known as the gal [Gal] in honor of Galileo. Note that the word for the unit is all lowercase, but the symbol is capitalized. The gal is an example of a Gaussian unit .

00 1 Gal = 1 cm/s 2  = 0.01 m/s 2 100 Gal = 100 cm/s 2  = 1 m/s 2 .

Split a gal into a thousand parts and you get a milligal [mGal].

1 mGal = 0.001 Gal  = 10 −5  m/s 2

Since Earth's gravity produces a surface acceleration of about 10 m/s 2 , a milligal is about 1 millionth of the value we're all used to.

1 g ≈ 10 m/s 2  = 1,000 Gal = 1,000,000 mGal

Measurements with this precision can be used to study changes in the Earth's crust, sea levels, ocean currents, polar ice, and groundwater. Push it a little bit further and it's even possible to measure changes in the distribution of mass in the atmosphere. Gravity is a weighty subject that will be discussed in more detail later in this book.

Don't confuse the phenomenon of acceleration due to gravity with the unit of a similar name. The quantity g has a value that depends on location and is approximately …

almost everywhere on the surface of the Earth. The unit g has the exact value of…

g =  9.80665 m/s 2

by definition.

They also use slightly different symbols. The defined unit uses the roman or upright g while the natural phenomenon that varies with location uses the italic or oblique g . Don't confuse g with g .

As mentioned earlier, the value of 9.8 m/s 2 with only two significant digits is valid for most of the surface of the Earth up to the altitude of commercial jet airliners, which is why it is used throughout this book. The value of 9.80665 m/s 2 with six significant digits is the so called standard acceleration due to gravity or standard gravity . It's a value that works for latitudes around 45° and altitudes not too far above sea level. It's approximately the value for the acceleration due to gravity in Paris, France — the hometown of the International Bureau of Weights and Measures . The original idea was to establish a standard value for gravity so that units of mass, weight, and pressure could be related — a set of definitions that are now obsolete. The Bureau chose to make this definition work for where their laboratory was located. The old unit definitions died out, but the value of standard gravity lives on. Now it's just an agreed upon value for making comparisons. It's a value close to what we experience in our everyday lives — just with way too much precision.

Some books recommend a compromise precision of 9.81 m/s 2 with three significant digits for calculations, but this book does not. At my location in New York City, the acceleration due to gravity is 9.80 m/s 2 . Rounding standard gravity to 9.81 m/s 2 is wrong for my location. The same is true all the way south to the equator where gravity is 9.780 m/s 2 at sea level — 9.81 m/s 2 is just too big. Head north of NYC and gravity gets closer and closer to 9.81 m/s 2 until eventually it is. This is great for Canadians in southern Quebec, but gravity keeps keeps increasing as you head further north. At the North Pole (and the South Pole too) gravity is a whopping 9.832 m/s 2 . The value 9.806 m/s 2 is midway between these two extremes, so it's sort of true to say that…

g  =  9.806 ± 0.026 m/s 2

This is not the same thing as an average, however. For that, use this value that someone else derived…

g  =  9.798 m/s 2

Here are my suggestions. Use the value of 9.8 m/s 2 with two significant digits for calculations on the surface of the Earth unless a value of gravity is otherwise specified. That seems reasonable. Use the value of 9.80665 m/s 2 with six significant digits only when you want to convert m/s 2 to g. That is the law.

The unit g is often used to measure the acceleration of a reference frame . This is technical language that will be elaborated upon later in another section of this book, but I will explain it with examples for now. As I write this, I'm sitting in front of my computer in my home office. Gravity is drawing my body down into my office chair, my arms toward the desk, and my fingers toward the keyboard. This is the normal 1 g (one gee) world we're all accustomed to. I could take a laptop computer with me to an amusement park, get on a roller coaster, and try to get some writing done there. Gravity works on a roller coaster just as it does at home, but since the roller coaster is accelerating up and down (not to mention side to side) the sensation of normal Earth gravity is lost. There will be times when I feel heavier than normal and times when I fell lighter than normal. These correspond to periods of more than one g and less than one g. I could also take my laptop with me on a trip to outer space. After a brief period of 2 or 3 g (two or three gees) accelerating away from the surface of the Earth, most space journeys are spent in conditions of apparent weightlessness or 0 g (zero gee). This happens not because gravity stops working (gravity has infinite range and is never repulsive), but because a spacecraft is an accelerating reference frame. As I said earlier, this concept will be discussed more thoroughly in a later section of this book.

• Physics Formulas

Free Fall Formula

Freefall as the term says, is a body falling freely because of the gravitational pull of our earth.

Imagine a body with velocity (v) is falling freely from a height (h) for time (t) seconds because of gravity (g).

Free Fall Formulas  are articulated as follows:

h = (1/2) gt 2

Freefall Related Solved Examples

Underneath are given questions on free fall which may be useful for you.

Problem 1:  Calculate  the  body  height  if  it  has  a  mass  of  2  kg  and  after  7 seconds  it  reaches  the  ground?

Given: Height h =? Time t = 7s We all are acquainted with the fact that free fall is independent of mass.

Hence, it is given as

h = 0.5 × 9.8 × (7) 2

h = 240.1 m

Problem  2: The cotton falls after 3 s and iron falls after 5 s. Which is moving with higher velocity? Answer:

The Velocity in free fall is autonomous of mass.

V (Velocity of iron) = gt = 9.8 m/s 2  × 5s = 49 m/s

V (Velocity of cotton) = gt = 9.8 m/s 2  × 3s = 29.4 m/s.

The Velocity of iron is more than cotton.

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Free Standing Fold Away Fall Arrest System

Fold Away Fall Arrest System for Fuel Transport Company

Problem: This Indiana Fuel Transport Company needed to provide an overhead fall protection anchorage for its service technicians as they worked on top of transport trailers in a very restricted service bay area.

Solution: The building, being of minimal structural capacity, was not able to support a traditional overhead rail type system. As a result, EFP installed a floor mounted free-standing system to address the customer's needs while refraining from using the building structure for support. This is a good example of the variety of system solutions available from EFP.

Engineered Fall Protection is a manufacturer and distributor for fall protection systems in Missouri, Illinois, Indiana, Kansas, Arkansas, Mississippi, Nebraska, Tennessee, Kentucky, Iowa, Oklahoma, Georgia, South Carolina, Alabama, and Florida.

• Gorbel Tether Track Rigid Lifelines
• Engineered Horizontal Lifelines
• Temporary Horizontal Lifelines
• Confined Space Rescue Equipment
• Railcar Fall Protection Systems
• Truck / Trailer Fall Protection
• Vertical Lifeline Systems
• Self Retracting Lifelines
• Safety Harnesses
• Safety Lanyards
• Heavy Duty Industrial Guardrail
• Temporary/Portable Guardrail
• Railcar Loading Platforms
• Truck Loading Platforms
• Engineered Platforms/Mezzanines
• Roof Anchor Posts
• Roof Fall Protection Kits
• Roof Ladder Systems
• Counterweight Anchor System
• Reusable Roof Anchors
• Steel Beam Anchors
• Vacuum Anchor Systems
• Tuff Built 4x4 Fall Arrest Anchor
• Malta Dynamics XSERIES Grabber
• Frame Style Systems
• Counterweight Jib Anchors
• Portable Fall Protection Platform
• Fall Protection Equipment Inspection
• Fall Protection Installation
• Fall Protection Site Assessment
• Fall Protection Training
• Fall Protection Case Studies
• Fall Protection Videos
• Ergonomic Lifting Solutions

(314) 492-4422