Question

A ball is thrown vertically upward from the top of a tower. Velocity at depth h from the point of projection is twice of the velocity at height h above the point of projection. Find the maximum height reached by the ball.

A
2h
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B
3h
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C
(53)h
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D
(43)h
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Solution

The correct option is C (53)h Here, |vB|=2|vA| By, applying 3rd equation of kinematics from A to B. u=vA v=vB v2−u2=2as ⇒v2B−v2A=2(−g)(−2h)4v2A−v2A=4gh3v2A=4ghv2A=4gh3....(1) Again using 3rd equation of kinematics from A to C: v2−u2=2as, here v=0, u=vA 0−v2A=2(−g)ss=v2A2g,from (1)s=4gh3×2g s=2h3 So maximum height reached by the ball =s+h=5h3

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