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

Explanation

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")=(u^(2))/(g)=100m`
If ball is projected vertically upwards `(theta=90^(@))` from ground then H attains maximum value.
`H_("max")=(u^(2))/(2g)=(R_("max"))/(2)`
`therefore` the height to which cricketer can through the ball is `=(R_("max"))/(2)=(100)/(2)=50m`.