A cricketer can throw a ball to a maximum horizontal distance of 100 m. 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 = 450. ThenRmax = u2gHere, Rmax = 100 m∴  u2g = 100 m  ......iAs v2-u2 =  2asHere v = 0 (At highest point velocity is zero)a = -g, s = H ∴  H = u22g = 1002 = 50 m  (Using (i))