Question

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

Answer 1

$\text{Step 1: Initial speed calculation [Refer Fig. 1]}$

Let $u$ be the initial velocity of the projectile.

Horizontal Range of the projectile is given by: $R=\frac{u^2\sin 2\theta }{g}$

For maximum range $\theta ={45}^{\circ }$

$R=\frac{u^2\sin {90}^o}{10}$

$\Rightarrow 100m=\frac{u^2}{10m/s^2}$

$\Rightarrow$ $u=\sqrt{1000}m/s$

$\text{Step 2: Maximum height calculation (H)[Refer Fig. 2]}$

The ball will achieve the maximum height$\left(H\right)$ when it is thrown vertically upwards

At topmost point, the velocity would be zero i.e $v=0$

Since acceleration $(a=-g)$ is constant, therefore applying equation of motion (Taking upward positive)

$v^2-u^2=2as$

$\Rightarrow$ $0^2-1000=-2\times 10\times H$

$\Rightarrow$ $H=50m$