Maximum horizontal distance, $R=100 \,m$ The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is $45^{\circ},$ i.e., $\theta=45^{\circ}$ The horizontal range for a projection velocity $v$, is given by the relation

$R=\frac{u^{2} \sin 2 \theta}{g}$

$100=\frac{u^{2}}{g} \sin 90^{\circ}$

$\frac{u^{2}}{g}=100$

The ball will achieve the maximum height when it is throwertically upward. For such motion, the final velocity $v$ is zero at the maximum height $H$ Acceleration, $a=-g$

Using the third equation of motion

$v^{2}-u^{2}=-2 g H$

$H=\frac{1}{2} \times \frac{u^{2}}{g}=\frac{1}{2} \times 100=50 \,m$