Free Fall (Physics): Definition, Formula, Problems & Solutions (W/ Examples)

​Free fall​ refers to situations in physics where the only force acting on an object is gravity.

The simplest examples occur when objects fall from a given height above the surface of the Earth straight downward – a one-dimensional problem. If the object is tossed upward or forcefully thrown straight downward, the example is still one-dimensional, but with a twist.

Projectile motion is a classic category of free-fall problems. In reality, of course, these events unfold in the three-dimensional world, but for introductory physics purposes, they are treated on paper (or on your screen) as two-dimensional: ​x​ for right and left (with right being positive), and ​y​ for up and down (with up being positive).

Free-fall examples therefore often have negative values for y-displacement.

It is perhaps counterintuitive that some free-fall problems qualify as such.

Keep in mind that the only criterion is that the only force acting on the object is gravity (usually Earth's gravity). Even if an object is launched into the sky with colossal initial force, at the moment the object is released and thereafter, the only force acting on it is gravity and it is now a projectile.

•Often, high-school and many college physics problems neglect air resistance,
though this always has at least a slight effect in reality; the exception is an event that unfolds in a vacuum. This is discussed in detail later.

A unique an interesting property of the acceleration due to gravity is that it is the same for all masses.

This was far from self-evident until the days of Galileo Galilei (1564-1642). That's because in reality gravity is not the only force acting as an object falls, and the effects of air resistance tend to cause lighter objects to accelerate more slowly – something we've all noticed when comparing the fall rate of a rock and a feather.

Galileo conducted ingenious experiments at the "leaning" Tower of Pisa, proving by dropping masses of different weights from the high top of the tower that gravitational acceleration is independent of mass.

Usually, you are looking to determine initial velocity (v_0y), final velocity (v_y) or how far something has fallen (y − y₀). Although Earth's gravitational acceleration is a constant 9.8 m/s², elsewhere (such as on the moon) the constant acceleration experienced by an object in free fall has a different value.

For free fall in one dimension (for example, an apple falling straight down from a tree), use the kinematic equations in the ​Kinematic Equations for Free-Falling Objects​ section. For a projectile-motion problem in two dimensions, use the kinematic equations in the section ​Projectile Motion and Coordinate Systems​.

•You can also use the conservation of energy principle, which states that ​the loss of potential energy (PE)​ during the fall ​equals the gain in kinetic energy (KE):​ –mg(y − y₀) = (1/2)mv_y². 

All of the foregoing can be reduced for present purposes to the following three equations. These are tailored for free fall, so that the "y" subscripts can be omitted. Assume that acceleration, per physics convention, equals −g (with the positive direction therefore upward).

•Note that v₀ and y₀ are initial values in any problem, not variables.

\(v=v_0-gt\text{ }\y=y_0+v_0t-\frac{1}{2}gt^2\text{ }\v^2=v_0^2-2g(y-y_0)\)

​Example 1:​ A strange birdlike animal is hovering in the air 10 m directly over your head, daring you to hit it with the rotten tomato you're holding. With what minimum initial velocity v₀ would you have to throw the tomato straight up in order to ensure that it reaches its squawking target?

What's happening physically is that the ball is coming to a stop owing to the force of gravity just as it reaches the required height, so here, v_y = v = 0.

First, list your known quantities: ​v =​ 0​, g =​ –9.8 m/s2​**, y − y₀ =**​ 10 m

Thus you can use the third of the equations above to solve:

\(0=v_0^2-2(9.8)(10)\text{ }\v_0^2=196\text{ }\v_0=14\text{ m/s}\)

This is about 31 miles an hour.

Projectile motion involves the motion of an object in (usually) two dimensions under the force of gravity. The behavior of the object in the x-direction and in the y-direction can be described separately in assembling the greater picture of the particle's motion. This means that "g" appears in most of the equations required to solve all projectile-motion problems, not merely those involving free fall.

The kinematic equations needed to solve basic projectile motion problems, which omit air resistance:

\(x=x_0+v_{0x}t\text{ }\v_y=v_{0y}-gt\text{ }\y-y_0=v_{0y}t-\frac{1}{2}gt^2\text{ }\v_y^2=v_{0y}^2-2g(y-y_0)\)

​Example 2:​ A daredevil decides to try to drive his "rocket car" across the gap between adjacent building rooftops. These are separated by 100 horizontal meters, and the roof of the "take-off" building is 30 m higher than the second (this almost 100 feet, or perhaps 8 to 10 "floors," i.e., levels).

Neglecting air resistance, how fast will he need to be going as he leaves the first rooftop to assure just reaching the second rooftop? Assume his vertical velocity is zero at the instant the car takes off.

Again, list your known quantities: (x – x₀) = 100m, (y – y₀) = –30m, v_0y = 0, g = –9.8 m/s².

Here, you take advantage of the fact that horizontal motion and vertical motion can be assessed independently. How long the car will take to free-fall (for purposes of y-motion) 30 m? The answer is given by y – y₀ = v_0yt − (1/2)gt^2.

Filling in the known quantities and solving for t:

\(−30 = (0)t − (1/2)(9.8)t^2\text{ }\30 = 4.9t^2\text{ }\t = 2.47\text{ s}\)

Now plug this value into x = x₀ + v_0xt :

\(100 = (v_{0x})(2.74)\implies v_{0x}=40.4\text{ m/s}\)

​**v_0x = 40.4 m/s (about 90 miles per hour).**​

This is perhaps possible, depending on the size of the roof, but all in all not a good idea outside of action-hero movies.

Air resistance plays a major, under-appreciated role in everyday events even when free fall is only part of the physical story. In 2018, a professional baseball player named Giancarlo Stanton hit a pitched ball hard enough to blast it away from home plate at a record 121.7 miles per hour.

The equation for the maximum horizontal distance a launched projectile can attain, or ​range equation​ (see Resources), is:

\(D=\frac{v_0^2\sin{2\theta}}{g}\)

Based on this, if Stanton had hit the ball at the theoretical ideal angle of 45 degrees (where sin 2θ is at its maximum value of 1), the ball would have traveled 978 feet! In reality, home runs almost never reach even 500 feet. Part if this is because a launch angle of 45 degrees for a batter is not ideal, as the pitch is coming in almost horizontally. But much of the difference is owed to the velocity-dampening effects of air resistance.

Free-fall physics problems aimed at less advanced students assume the absence of air resistance because this factor would introduce another force that can slow or decelerate objects and would need to be mathematically accounted for. This is a task best reserved for advanced courses, but it bears discussion here nonetheless.

In the real world, the Earth's atmosphere provides some resistance to an object in free fall. Particles in the air collide with the falling object, which results in transforming some of its kinetic energy into thermal energy. Since energy is conserved in general, this results in "less motion" or a more slowly increasing downward velocity.