Question

A cricketer can throw a ball to a maximum horizontal distance of $100\text{m}$. How much high above the ground can the cricketer throw the same ball?

Answer 1

Step 1: Given

The distance the cricketer can throw the ball horizontally, $R=100\text{m}$

Step 2: Formulas used

For a projectile motion, we know that the range (horizontal distance covered) is given as,
$R=\frac{V^2\sin \left(2\theta \right)}{g}$
where $V$ is the initial velocity, $\theta$ is the angle at which object was thrown at and $g$ is the acceleration due to gravity.

And the maximum height reached is given as,
$H=\frac{V^2{\sin }^2\left(\theta \right)}{2g}$
where the symbols mean the same as described earlier.

Step 3: Calculating height

We know that the maximum horizontal distance is covered when $\theta =45°$.

So we have,
$\begin{array}{ccc}& R=\frac{V^2\sin \left(2\theta \right)}{g}& \\ \Rightarrow & 100=\frac{V^2\sin \left(2\times {\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{g}& \left[\because 45°=\frac{\mathrm{\pi }}{4}\text{rad}\right]\\ \Rightarrow & \frac{V^2}{g}=100& \left[\because \sin \left(\frac{\mathrm{\pi }}{2}\right)=1\right]\dots \left(1\right)\end{array}$

We also have that,
$\begin{array}{ccc}& H=\frac{V^2{\sin }^2\left(\theta \right)}{2g}& \\ \Rightarrow & H=\frac{100{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{2}& \left[\because \left(1\right)\right]\\ \Rightarrow & H=50{\left(\frac{1}{\sqrt{2}}\right)}^2& \left[\because \sin \left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}\right]\\ \Rightarrow & H=25\text{m}& \end{array}$

Therefore, the maximum height the cricketer can throw the ball is $25\text{m}$.