Question

A cricketer can throw a ball to a maximum horizontal distance of $100\,m$. With the same speed how much high above the ground can the cricketer throw the same ball?

• A. $50\,m$
• B. $100\,m$
• C. $150\,m$
• D. $200\,m$

Answer 1

The Correct Option is A

Let $u$ be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, $\theta=45^{?}$. Then $R_{max}=\frac{u^{2}}{g}$ Here, $R_{max}=100\,m\,$ $\therefore \frac{u^{2}}{g}=100\,m\,...\left(i\right)$ As $v^{2}-u^{2}=2as$ Here, $v = 0$ (At highest point velocity is zero)

$a=-g$, $s=H$ $\therefore H=\frac{u^{2}}{2g}=\frac{100}{2}$ $=50\,m$ (Using (i))