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 ?

Explanation

Let u be the velocity of projection of the ball. The ball will cover maximum horizontaal distance when angle of projection with horizontal , `theta = 45^(@)`. Then
`R_(max) = (u^(2))/(g) = 100 ` m
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 throw the ball is `= (R_(max))/(2) = (100)/(2) = 50 `m