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Q.
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?
Motion in a Plane
Solution:
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))
