Question

a stone projected vertically up with velocity v from the top of a tower reaches the ground with velocity 2v. The height of the tower is​

Answer 1

Answer:

Take height of tower as - H

Initial energy = 1/2 mu^2 + mgH ( K.E + P.E)

Final energy when it hits ground = 1/2 m (3u)^2. ( only K.E) since its on ground (H=0)

Now consider there is no air friction loss. So initial energy equal to final energy.

1/2 mu^2 + mgH = 1/2 m (9mu^2)

Solve and u will get - H = 4u^2/g

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