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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
v2u2=2as
v2Bv2A=2(g)(2h)4v2Av2A=4gh3v2A=4ghv2A=4gh3....(1)

Again using 3rd equation of kinematics from A to C:
v2u2=2as, here v=0, u=vA
0v2A=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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