#### 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?

#### Solution 1

R_"max" = 100 m

`Since `R_"max" = v^2/g => 100 = v^2/g`

Using equation of motion

`v^2 - u^2 = 2as`

Here, v = 0, a = -g, `s = R_"max"` = 100 m

`:.(0)^2 -u^2 = 2(-g) xx s`

`=> s = 1/2 u^2/g`

Since u = v

`:.s = 1/2 v^2/g = 1/2 xx 100 = 50 m`

#### Solution 2

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., *θ* = 45°.

The horizontal range for a projection velocity *v*, is given by the relation:

`R = (u^2sin2theta)/g`

`100 = u^2/g sin 90^@`

`u^2/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, *a = –g*

Using the third equation of motion

`v^2 - u^2 = -2gH`

`H=1/2xxu^2/g = 1/2xx100 = 50 m`