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

Explanation

Let (u) be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle or projection with whorizontal, ` theta = 45^@. Then, ` R_(max) = u^2 //g` …(i0
In order to study to the motion of the ball along vertcal direction, consider a point on the surface of earth as the origin and vertical upward direction as the positive direction of (Y) axis . Taking motion of the ball along vertical upward direction we have ` u_y = u, a_y =- g. v_y =0, t= ?, y+0 =0 y=?`
Also , ` y=y_0 = u_y t = 1/2 a_y t^2`
:. ` y=0 + u ( u /g) + 1/2 (- g ) u^2 /g - 1/2 u^2 /g = 1/2 = ( 1000/2 = 50 m` [ from (i) ]` .