Answer :

Given:

Maximum horizontal distance travelled by the ball=Range =R= 100m

Explanation: We know that range is maximum if the angle of projection is 45˚.

Figure showing projectile motion

(ϕ = 45˚)

u = 31.304 m/s

Height to which the ball is thrown = H

Since the final velocity of the ball is zero i.e. v= 0

From the relation v^{2}-u^{2}=2as

⇒ 0 - 31.304^{2} = 2×9.8×H (s=H)

⇒ H= 50m

The cricketer can throw the ball to a height of 50m above the ground.

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