Free Fall

Free fall is the vertical movement that occurs when an object is subject only to the effect of gravitational force.

When a basketball is dropped towards the ground, every second it travels a greater distance because it is subjected to a gravitational force. Its acceleration is equal to the Earth's gravitational acceleration.

Earth's gravitational acceleration is not the same as the gravitational acceleration of other celestial bodies in the solar system. The magnitude of the acceleration depends, among other things, on the mass of the celestial body. The Moon, which is 81 times smaller than Earth, has a gravitational acceleration that is six times smaller than that of Earth, that is |{g} = 1.6 \: \text {m/s}^{2}.|

A ball is tossed upwards at a speed of |15.0\ \text {m/s}.| The ball is launched from a height of |1.2\ \text{m}| above the ground. What will be the maximum height reached by the ball?

In this type of problem, remember that when the ball reaches its maximum height, the speed is always equal to

\begin{align}a &= g = -9.8 \: \text{m/s}^2, & x_{i} &= 1.2 \: \text{m}, \\v_{i} &= 15.0 \: \text{m/s}, & v_{f} &= 0 \: \text{m/s}, \\x_{f} &= \: ?\end{align}
Using one of the UARM equations, it's possible to find the final position of the ball.

\begin{align}{v_{f}}^2 &= {v_{i}}^2 + 2 \cdot a \cdot \triangle x \quad \Rightarrow \quad \triangle x = \frac{{v_{f}}^2 - {v_{i}}^2}{2 \cdot a} \\&= \frac{(0 \: \text{m/s})^2 - (15 \: \text{m/s})^2}{2 \cdot (-9.8 \: \text{m/s}^2)} \\&= \frac{-225}{-19.6} \\&= 11.5 \: \text{m}\end{align} \begin{align} \triangle x = x_f - x_i \quad \Rightarrow \quad x_f &= \triangle x + x_i \\ &= 11.5 \: \text {m} + 1.2 \: \text {m} \\ &=12.7 \: \text {m}  \end{align}

A ball is thrown upwards from the roof of a building measuring |120 \: \text {m}|, at a speed of |4.0 \: \text {m/s}.| How long will it take for the ball to reach the ground?

First, consider the upward movement the ball will make.
||\begin{align}a &= g = -9.8 \:\text{m/s}^2 &x_{i} &= 12 \: \text{m} \\ v_i &= 4.0 \:\text{m/s} &v_{f} &= 0 \: \text{m/s}\\
\triangle t &= ? \end{align}||
Using the UARM equations, we can find the time required for the ball to reach the highest point..
||\begin{align} {v_{f}}={v_{i}}+ a \cdot \triangle t
\quad \Rightarrow \quad
\triangle t &=\frac {{v_{f}} -{v_{i}}}{a} \\
&= \frac {{0 \: \text {m/s}} -{4 \: \text {m/s}}}{-9.8 \: \text {m/s}^2}\\
&= 0.41 \: \text{s} \end{align}||
The next step is to determine the maximum height reached by the ball.
||\begin{align} {v_{f}}^2={v_{i}}^2+2 \cdot a \cdot \triangle x
\quad \Rightarrow \quad
\triangle x &=\frac {{v_{f}}^2 -{v_{i}}^2}{2 \cdot a} \\
&= \frac {{(0 \: \text {m/s})}^2 -{(4 \: \text {m/s})}^2}{2 \cdot -9.8 \: \text {m/s}^2}\\
&= 0.8 \: \text{m} \end{align}||
||\begin{align} \triangle x = x_f - x_i
\quad \Rightarrow \quad
x_f &= \triangle x + x_i \\
&= 12 \: \text {m} + 0.8 \: \text {m}\\
&= 12.8 \: \text{m} \end{align}||
For the second part, consider that the object is in free fall from its highest point until it reaches the ground.
||\begin{align}a &= g = -9.8 \:\text{m/s}^2 &\triangle x &= -12.8 \: \text{m} \\ v_i &= 0 \:\text{m/s}
&\triangle t &= ? \end{align}||
Using one of the UARM equations, we can find the time required for the object to reach the ground.
||\begin{align} \triangle x= v_{i} \cdot \triangle t + \frac{1}{2} \cdot a \cdot {\triangle t}^{2}
\quad \Rightarrow \quad
\triangle x&= 0 \: \text {m/s} \cdot \triangle t + \frac{1}{2} \cdot a \cdot {\triangle t}^{2}\\
\triangle x&= \frac{1}{2} \cdot a \cdot {\triangle t}^{2}\\
\triangle t&= \sqrt{\frac {2 \cdot \triangle x}{a}} \\
&= \sqrt{\frac {2 \cdot -12.8 \: \text {m}}{-9.8 \: \text {m/s}^2}} \\
&= 1.62 \: \text{s} \end{align}||
Given that the upward movement lasted |0.41 \: \text{s}| and the downward movement had a duration of |1.62 \: \text{s}|, the time required for the ball to reach the ground is |0.41 \: \text{s}+1.62 \: \text{s}=2.03 \: \text{s}.|

It is also possible to start to study the motion at maximum height and record the position until the object reaches the ground. It is very important to establish the position of the reference system before graphing.