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?

Given,
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)
= (1/2)(100)
= 50
Hence, the maximum height = 50 m

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