Question

A ball is thrown vertically upwards from the top of a tower. Velocity at a point 'h'. vertically below the point of projection is twice the downward velocity at a point "h m vertically above the point of projection. The maximum height reached by the ball above the top of the tower is​

Answer 1

Explanation:

Let 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 Kαv

2

. Thus, by energy conservation,

K+mgh=4K−mgh

K=

3

2mgh

Also, let the maximum height reached above tower be H. Then, again by conserving energy,

K+mgh=

3

2mgh

+mgh=mgH

Hence, H=

3

5h

$.$

Hope it helps you mate:)

Answer 2

Answer:

5h/3

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