Question

# A ball is thrown vertically upwards from the top of a tower. Velocity at a point â€²hâ€² m vertically below the point of projection is twice the downward velocity at a point â€²hâ€² m vertically above the point of projection. The maximum height reached by the ball above the top of the tower is:

A
4h3
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B
5h3
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C
3h
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D
2h
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Solution

## The correct option is A 5h3Let the velocity at height h above the tower be v and kinetic energy at this point be K. Then the velocity at height h below the tower will be 2v. Kinetic energy at this point will then be 4K since Kαv2. Thus, by energy conservation,K+mgh=4K−mghK=2mgh3Also, let the maximum height reached above tower be H. Then, again by conserving energy,K+mgh=2mgh3+mgh=mgHHence, H=5h3

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