Horizontal distance = 100 m
Let u be the velocity of projection of the ball.
We know that maximum horizontal range at the angle of projection with horizontal, θ = 45°.
Maximum Range = u2/g
100 = u2/g
Using the relation, v2 – u2 = 2as
Here, v = 0, a = -g and s = R(max) = 100 m
So, 0 – u2 = -2gs
s = (1/2)(u2/g)
Hence, the maximum height = 50 m