Analysing Graphs of Projectile Motion

The graph below (not drawn to scale) shows the motion of a tennis ball that was thrown vertically upwards from an open window some distance from the ground. It takes the ball \(\text{0.2}\) \(\text{s}\) to reach its highest point before falling back to the ground. Study the graph given and calculate:

1. how high the window is above the ground.

2. the time it takes the ball to reach the maximum height.

3. the initial velocity of the ball.

4. the maximum height that the ball reaches.

5. the final velocity of the ball when it reaches the ground.

Step 1: Find the height of the window

The initial position of the ball will tell us how high the window is. From the y-axis on the graph we can see that the ball is \(\text{4}\) \(\text{m}\) from the ground.

The window is therefore \(\text{4}\) \(\text{m}\) above the ground.

Step 2: Find the time taken to reach the maximum height

The maximum height is where the position vs time graph reaches its maximum position. This is when t = \(\text{0.2}\) \(\text{s}\).

It takes the ball \(\text{0.2}\) \(\text{s}\) to reach the maximum height.

Step 3: Find the initial velocity ( \(\vec{v}_{i}\)) of the ball

To find the initial velocity we only look at the first part of the motion of the ball. That is from when the ball is released until it reaches its maximum height. We have the following for this: In this case, let's choose upwards as positive.

\begin{align*} t& = \text{0,2} \text{s} \\ \vec{g}& = \text{9,8}\text{ m·s $^{-2}$}\\ \vec{v}_{f}& = 0 \text{m}·{\text{s}}^{-1} \text{(because the ball stops)} \end{align*}

To calculate the initial velocity of the ball ( \(\vec{v}_{i}\)), we use:

\begin{align*} \vec{v}_{f}& = \vec{v}_{i}+\vec{g}t \\ 0& = \vec{v}_{i}+\left( -\text{9,8}\right)\left( \text{0,2}\right) \\ {v}_{i}& = \text{1,96} \text{m}·{\text{s}}^{-1} \end{align*}

The initial velocity of the ball is \(\text{1.96}\) \(\text{m·s $^{-1}$}\)upwards.

Step 4: Find the maximum height of the ball

To find the maximum height we look at the initial motion of the ball. We have the following:

\begin{align*} t& = \text{0,2} \text{s} \\ \vec{g}& = \text{9,8}\text{ m·s $^{-2}$}\\ \vec{v}_{f}& = \text{0} \text{m}·{\text{s}}^{-1}~\text{(because the ball stops)} \\ \vec{v}_{i}& = \text{+1,96} \text{m}·{\text{s}}^{-1}\text{(calculated above)} \end{align*}

To calculate the displacement from the window to the maximum height (\(\Delta x\)) we use:

\begin{align*} \Delta \vec{x}& = \vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2} \\ \Delta \vec{x}& = \left( \text{1,96}\right)\left( \text{0,2}\right)+\frac{1}{2}\left( -\text{9,8}\right){\left( \text{0,2}\right)}^{2} \\ \Delta \vec{x}& = \text{0,196} \text{m} \end{align*}

The maximum height of the ball is \(\left(\text{4}+\text{0.196}\right)= \text{4.196} \text{m}\) above the ground.

Step 5: Find the final velocity ( \(\vec{v}_{f}\)) of the ball

To find the final velocity of the ball we look at the second part of the motion. For this we have:

\begin{align*} \Delta \vec{x}& = -\text{4,196} \text{m}~\text{(because upwards is positive)} \\ \vec{g}& = -\text{9,8}\text{ m·s $^{-2}$}\\ \vec{v}_{i}& = \text{0} \text{m}·{\text{s}}^{-1} \end{align*}

We can use \(\vec{v}^2_{f}=\vec{v}^2_{i}+2\vec{g}\Delta\vec{x}\) to calculate the final velocity of the ball.

\begin{align*} \vec{v}^2_{f}=\vec{v}^2_{i}+2\vec{g}\Delta\vec{x} \\ \vec{v}^2_{f}& = {\left(0\right)}^{2}+2\left( -\text{9,8}\right)\left( -\text{4,196}\right) \\ \vec{v}^2_{f}& = \text{82,2416}\\ \vec{v}_{f}& = \text{9,0687}\text{ m·s$^{-1}$}~\text{downwards} \end{align*}

The final velocity of the ball is \(\text{9.07}\) \(\text{m·s $^{-1}$}\) downwards.