Question

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

Answer 1

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°, i.e., $\theta$ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
$\text R$ = $\frac{\text u^2 \sin 2\theta}{\text g}$

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

$\frac{\text u^2}{\text g}$ = 100 ...(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, $\text a$ = $-\text g$

Using the third equation of motion:
$\text v^2-\text u^2$ = $-2\text{gH}$

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