Question

A ball of mass m is released from A at a height H above the ground Find the kinetic energy of ball when it passes B which is at a distance H 12 above from the ground

Answer 1

Total mechanical energy at height, H

EH=mgH

Let vh be velocity of the ball at height h(=43H)

∴ Total mechanical energy at height h,

Eh=mgh+21mvh2

According to law of conservation of mechanical energy,

EH=Eh;mgH=mgh+21mvh2

vh2=2g(H−h)

Required ratio of kinetic energy to potential energy at height h is

VhKh=mgh21mvh2=mgh21m2g(H−h)=(hH−1)=31.