A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively

Solution: Answer (4)

Let required height of body is y.

When body from rest falls through height (S – y)

Then under constant acceleration

v² = 0² + 2g(S – y)

v = {2g(S-y)}^1/2 …(1)

When body is at height y above ground. The potential energy of body of mass m

U = mgy

As per given condition kinetic energy, K = 3U