Q.
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?
Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, θ=45°. Then Rmax=gu2
Here, Rmax=100m ∴gu2=100m...(i)
As v2−u2=2as
Here, v=0 (At highest point velocity is zero) a=−g, s=H ∴H=2gu2=2100 =50m (Using (i))