Question

A cricketer can throw a ball to a maximum distance of 100 m. With the same speed how high above the ground can the cricketer throw the same ball.

Answer 1

Maximum range of a projectile thrown at velocity v is given by

$R_{max}= \frac{v^2}{g}$

100 = v²/g
⇒ 100 = v²/9.8
⇒ v = √100×9.8
⇒ v = 31.3 m/s

With same speed, if we want to find out the maximum height, throw it vertically upward at same speed. 
Maximum height = v²/2g = 31.3²/(2×9.8) = 50 m