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