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., θ = 45°.

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

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

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

`v^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`

= `(0)^2 - (sqrt(100)g)^2 = 2(-g)H`

`H=(100g)/(2g)=50  "m"`