Chapter 5. Free Fall

There are several real-life situations that fall under the category of free-falling motion:

Regardless of the situation, free-fall motion is described by the same formulas as accelerated motion. That is because:

1. On its way upward, an object will decelerate with an acceleration equal to [latex]g = 9.8 \, \text{m/s}^2[/latex] (the acceleration due to gravity) until it reaches the maximum height. At the maximum height, the instantaneous velocity is zero.
2. On its way downward, an object will accelerate with an acceleration equal to [latex]g = 9.8 \, \text{m/s}^2[/latex] (the acceleration due to gravity).

It follows that all the equations used for accelerated motion apply to the free-fall motion with the note that acceleration is always [latex]g = 9.8 \, \text{m/s}^2[/latex] near the surface of the Earth. On another planet, the value of acceleration due to gravity will change. 

Example

On the surface of the Moon, the acceleration due to gravity is approximately 6 times less than that on Earth. 

Throwing an object upwards with an initial velocity [latex]v_i[/latex]

Notations and analysis of the motion 

Notations: 

[latex]v_{i, f}[/latex] represent the object’s initial velocity and final velocity respectively;

[latex]v[/latex] represents the velocity at the maximum height (equal to zero);

[latex]h_{max}[/latex] represents the maximum height reached by the object;

[latex]t_{up, down}[/latex] represent the time needed to reach the maximum height and the time to come back down from the maximum height respectively;

[latex]g[/latex] represents acceleration due to gravity.

Analysis of the motion 

From point 1 to point 2 (see Figure 5.2):

To calculate the maximum height, we use the following equation: 

[latex]v_f^2 = v_i^2 + 2aS[/latex]

which becomes: 

[latex]v_{f}^2 = v_{i}^2 - 2gh_{\text{max}}[/latex]   (1)

The acceleration is negative because the object loses speed when going upward (decelerated motion).

 From equation (1), the maximum height becomes:

[latex]h_{\text{max}} = \frac{v_{i}^2}{2g}[/latex]   (2)

From point 2 to point 3 (see Figure 5.2):

On its way down, the object will come back down from the maximum height, where the velocity is zero, to the same level where it started.

We can express the maximum height using the same formula above:

[latex]v_f^2 = v_i^2 + 2aS[/latex]

which becomes:

[latex]v_{f}^2 = v_{i}^2 + 2gh_{\text{max}}[/latex]   (3)

In this case [latex]v_i = 0[/latex], and the acceleration is positive, which means:

[latex]h_{\text{max}} = \frac{v_f^2}{2g}[/latex]   (4)

Since expressions (2) and (4) represent the same quantity, they are equal.

Conclusion: The initial velocity (at the time of the throw) is equal to final velocity (at the time the object reaches the same level on its way down).

Using this concept, it can be shown that the time to reach the maximum height is equal to the time to come back from the maximum height to the same level.

So, the following equations can be used in solving problems of free fall:

[latex]v_i = v_f[/latex]   (5)

[latex]t_{\text{up}} = t_{\text{down}}[/latex]   (6)

An important note! The equations above are true ONLY for situations in which the object returns to the same level from which it started.

Chapter equations

[latex]v_{f}^2 = v_{i}^2 - 2gh_{\text{max}}[/latex]   (1)

[latex]h_{\text{max}} = \frac{v_{i}^2}{2g}[/latex]   (2)

[latex]v_{f}^2 = v_{i}^2 + 2gh_{\text{max}}[/latex]   (3)

[latex]h_{\text{max}} = \frac{v_f^2}{2g}[/latex]   (4)

[latex]v_i = v_f[/latex]   (5)

[latex]t_{\text{up}} = t_{\text{down}}[/latex]   (6)

Try it! 

Practice