Height of a Projectile Example

A cricketer hits a cricket ball so that it goes vertically upwards. If the ball takes \(\text{10}\) \(\text{s}\) to return to the initial height, determine its maximum height above the initial position.

Step 1: Identify what is required and what is given

We need to find how high the ball goes. We know that it takes \(\text{10}\) \(\text{s}\) to go up and down. We do not know what the initial velocity (\(\vec{v}_{i}\)) of the ball is.

After the batsman strikes the ball it goes directly upwards and the only force that will be acting on the ball will be the gravitational force. This information is represented in the photograph below:

Step 2: Determine how to approach the problem

For a problem like this it is useful to divide the motion into two parts. The first part of the motion consists of the upward motion of the ball with an initial velocity (\(\vec{v}_{i}\)) and a final velocity (\(\vec{v}_{f}=0\)) at the maximum height. The second part of the motion consists of the downward motion of the ball with an initial velocity (\(\vec{v}_{i}=0\)) and final velocity (\(\vec{v}_{f}\)) which is not yet known.

Choose down as positive. We know that at the maximum height, the velocity of the ball is \(\text{0}\) \(\text{m·s $^{-1}$}\). We also know that the ball takes the same time to reach its maximum height as it takes to travel from its maximum height to the initial position, due to time symmetry. The time taken is half the total time. Therefore, we have the following information for the second (downward) part of the motion of the ball:

• \(t=\text{5}\text{ s}\) (half of the total time)
• \(\vec{v}_{\mathrm{top}}=\text{0}\text{ m·s$^{-1}$}\)
• \(\vec{g}= \text{9.8}\text{ m·s $^{-2}$}\) downwards
• \(\Delta \vec{x}=~?\)

Step 3: Find an appropriate equation to use

We do not know the final velocity of the ball coming down. We need to choose an equation that does not have \(\vec{v}_{f}\) in it. We can use the following equation to solve for \(\Delta \vec{x}\):

\[\Delta \vec{x}=\vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2}\]

Step 4: Substitute values and find the height

\begin{align*} \Delta \vec{x}& = \left( \text{0}\right)\left( \text{5}\right)+\frac{ \text{1}}{ \text{2}}\left( \text{9,8}\right){\left( \text{5}\right)}^{2} \\ \Delta \vec{x}& = \text{122,5}\text{ m}~\text{downwards}\end{align*}

In the second motion, the displacement of the ball is \(\text{122.5}\) \(\text{m}\) downwards. This means that the height was \(h = \text{122.5}\text{ m}\).

Step 5: Write the final answer

The ball reaches a maximum height of \(\text{122.5}\) \(\text{m}\).