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:
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 v2-u2=2as
⇒ 0 - 31.3042 = 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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