Every year in the US, over \( 350,000 \) people skydive. When skydiving, jumpers jump out of airplanes \( 14,000\,\mathrm{ft} \) above the ground. Once in the air, jumpers travel about \( 305\,\mathrm{m} \) every \( 5\,\mathrm{s}, \) all while experiencing an effect, known to avid jumpers, as "relative wind." Relative wind allows jumpers to deflect the rushing air over their bodies and provides them with very accurate control of what they do in the air from flips, spins, etc. Eventually, jumpers will reach a maximum speed of \( 53.6\,\mathrm{\frac{m}{s}} \) before having to pull their parachutes. Therefore, skydiving is a prime example of free-falling. Now if you have experienced this firsthand, you may be familiar with concepts such as air resistance and terminal velocity. However, if not, let this article introduce the concepts free-falling through definitions and examples.
Figure 1: Skydiving is a prime example of free fall motion.
Free Falling
Below we define and discuss free falling and its characteristics.
Free falling is the linear motion of an object in which only the force of gravity is acting on the object.
Linear motion is a one-dimensional motion.
When objects are in free fall, these objects are assumed to fall within a vacuum. As a result, this motion is defined by two characteristics:
- Objects do not experience air resistance.
- Objects accelerate towards earth at \( 9.8\,\mathrm{\frac{m}{s^2}}. \)
Air resistance, also known as drag, is a force that acts on an object in the opposite direction compared to the force of an object's weight.
Weight of a Free-Falling Object
When an object is in free fall, the gravitational force acting on it is said to be equal to the weight of the object. The formula for weight is given by the equation $$\begin{align}W&=mg\\W&=F\\\end{align},$$ where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(F) is the force of weight.
Equations of a Free-Falling Object
Variables associated with the motion of free falling can be calculated through the use of kinematics.
Kinematics refers to the study of motion without considering the forces at play.
In the field of kinematics, there are four corresponding variables of motion. These variables are
- Displacement
- Velocity
- Acceleration
- Time
Note that velocity, acceleration, and displacement are all vector quantities meaning they have magnitude and direction.
Thus, let us define each variable starting with displacement.
Displacement is the product of velocity and time.
The mathematical formula corresponding to this definition is $$\Delta{y}=\Delta{v}\,\Delta{t}$$ where \( \Delta{v} \) is velocity and \( t \) is time. Displacement has a SI unit of \(\mathrm{m}\).
Velocity is an object's change in displacement with respect to time.
The mathematical formula corresponding to this definition is
$$v=\frac{\Delta{y}}{\Delta{t}}$$ where \( \Delta{y} \) is displacement and \( t \) is time. Velocity has a SI unit of \( \mathrm{\frac{m}{s}} \).
Acceleration is an object's change in velocity with respect to time.
The mathematical formula corresponding to this definition is
$$a=\frac{\Delta{v}}{\Delta{t}}$$ where \( v \) is velocity and \( t \) is time. Acceleration has a SI unit of \( \mathrm{\frac{m}{s^2}} \).
Time does not change relative to an object's type of motion and is measured in \( s. \)
Relationship between Variables
Consequently, three equations describe the relationship between these variables. These equations are the kinematic equations of motion.
The velocity equation,
$$v=v_{o} + g{t}.$$
The displacement equation,
$$\Delta{y} =v_o{t}+\frac{1}{2}{g}t^2.$$
The velocity squared equation,
$$v^2=v_{o}^2 +2g\Delta{y}.$$
The average velocity equation,
$$v_{avg}=\frac{v+v_o}{2}.$$
where \( v \) is final velocity, \( v_o \) is the initial velocity, \( g \) is acceleration due to gravity, \( t \) is time, and \( \Delta{y} \) is displacement.
These kinematic equations only apply when acceleration is constant.
Note that the gravitational constant, \( g \), has replaced acceleration, \( a \), in these equations and \( \Delta{x} \) has been replaced by \( \Delta{y} \) because we are working along the y-axis.
Acceleration of a Free-Falling Object
The acceleration of an object in free fall is equal to gravity. This can be seen by rearranging Newton's second law of motion, \( F=ma \), and solving for acceleration.
$$\begin{align}F&=ma\\a&=\frac{F}{m}=\frac{W}{m}=\frac{mg}{m}=g\\a&=g.\\\end{align}$$
Objects in free fall are accelerating downward toward earth, therefore, we must be aware that acceleration is negative.
This implies that objects will accelerate at the same rate regardless of their shape, size, or mass. However, we must understand that this neglects air resistance. For example, if air resistance was not neglected, a ball and sheet of paper in free fall would not accelerate at the same rate. The ball would accelerate faster and reach the ground first because it has a larger mass and larger momentum in comparison to the sheet of paper.
Speed and Velocity of a Free-Falling Object
Objects in free-fall accelerate downward toward the Earth's surface. This downward motion indicates negative acceleration. As acceleration tells us how fast velocity changes every second, the downward direction also implies that velocity will be negative. This can be explained by the fact that acceleration and velocity are vector quantities, with magnitude and direction. As the direction is downward, a negative value is implied. However, although velocity is negative, speed will be positive. As speed is a scalar quantity, with only magnitude, it is not affected by direction. Therefore, for an object in free fall, its velocity becomes more negative every second, while its speed becomes more positive every second.
To calculate the velocity and speed of an object in free fall, the kinematic equations, mentioned earlier, can be used.
Position Function for Free Falling Objects
To calculate an object's position, during free fall, with respect to time, the equation \( \Delta{y} =v_o{t}+\frac{1}{2}{g}t^2 \) can be used. However, we can note that objects in free fall start from rest, indicating the object's initial velocity is zero. As a result, the equation can be simplified to \( \Delta{y}=\frac{1}{2}gt^2. \) The equation \( v^2=v_{o}^2 +2g\Delta{y} \) also can be used to determine position.
Free Falling Object in Equilibrium
When an object is initially released from rest, the only force acting on it is its weight which is given by \( W=mg. \) As the object accelerates downward, it begins to experience the effects of air resistance. Air resistance opposes the force of weight, but the object's weight will be greater allowing the object to continue accelerating. However, at a certain point, the object will reach its terminal velocity, its maximum speed, where air resistance becomes equal to weight. As a result, the object is then considered to be in equilibrium because it is no longer accelerating. This can be seen in the free-body diagram below.
Free body diagram of an object falling at terminal velocity
Let us use the free-body diagram below as a visual representation of our discussion of free fall and terminal velocity.
Figure 2: A free-body diagram demonstrating an object in free fall.
Examples of how to solve objects in free fall problems
To solve free fall problems, kinematic equations can be applied to calculate an object's acceleration, velocity, speed, and position. As we have discussed free-falling objects, their motion, and the forces affecting them, let us work through some examples to gain a better understanding of each concept. Note that before solving a problem, we must always remember these simple steps:
- Read the problem and identify all variables given within the problem.
- Determine what the problem is asking and what formulas are needed.
- Apply the necessary formulas and solve the problem.
- Draw a picture if necessary to provide a visual aid.
Examples
Let us apply our new knowledge of kinematics and free fall motion to the following two examples.
A ball is dropped out the window of a building. What is the displacement of the ball after \( 2\,\mathrm{s}? \)
Figure 3: Calculating the displacement of a ball in free fall.
Based on the problem, we are given the following:
As a result, we can identify and use the equation, \( \Delta{y}=\frac{1}{2}gt^2, \) to solve this problem. Therefore, our calculations are:
$$\begin{align}\Delta{y}&=\frac{1}{2}gt^2\\\Delta{y}&=\frac{1}{2}(9.8\,\mathrm{\frac{m}{s^2}})(2\,\mathrm{s})^2\\\Delta{y}&=19.6\,\mathrm{m}\\\end{align}$$
The displacement of the ball after \( 2\,\mathrm{s} \) is \( 19.6\,\mathrm{m}. \)
Now, after solving for displacement, let's solve for a different variable in the example below.
A rock, initially at rest, is dropped off of a \( 85\,\mathrm{m} \) cliff. What is the final velocity of the rock?
Figure 4: Calculating the final velocity of a rock in free fall.
Based on the problem, we are given the following:
- initial velocity
- displacement
As a result, we can identify and use the equation, \( v^2=v_{o}^2 +2g\Delta{y} \) to solve this problem. Therefore, our calculations are:
$$\begin{align} v^2&=v_{o}^2 +2g\Delta{y}\\ v^2&=0^2 +2\left(9.8\,\mathrm{\frac{m}{s^2}}\right)(85\,\mathrm{m})\\v^2&=1666\,\mathrm{\frac{m}{s}}\\v&=40.8\,\mathrm{\frac{m}{s}}\\\end{align}$$
The final velocity of the rock is \( 40.8\,\mathrm{\frac{m}{s}}. \)
Free Falling Object - Key takeaways
- Free falling is the linear motion of an object where only the force of gravity is acting on the object.
- Objects in free fall are assumed to be falling within a vacuum meaning objects do not experience air resistance and accelerate towards earth at \( 9.8\,\mathrm{\frac{m}{s^2}}. \)
- The gravitational force acting on an object in free fall is equal to the weight of the object, given by the equation, \( W=mg. \)
- The acceleration of an object in free fall is equal to gravity.
- Objects accelerate at the same rate regardless of their shape, size, or mass.
- Speed, velocity, and position of an object in free fall can be calculated by the linear motion kinematic equations.
- Acceleration and velocity, vector quantities, are negative while speed, a scalar quantity, is positive.
- A free-falling object is in equilibrium when air resistance becomes equal to the object's weight.
References
- Figure 1: Skydiving (https://www.pexels.com/photo/parachuters-on-air-70361/) by Pexel (www.pexel.com) licensed by CC0 1.0 Universal.
Explanations 
Exams 
Magazine 
