A cricketer can throw a ball to a maximum horizontal distance of 100 m.

solution:

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:

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

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

$\frac{u^{2}}{g}=100$ $\ldots(i)$

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.

Acceleration, $a=-g$

Acceleration, a = –g

Using the third equation of motion:

$v^{2}-u^{2}=-2 \mathrm{~g} H$

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