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 : 

\(\frac{S}{4}\), \(\sqrt{\frac{3 g S}{2}}\)

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 = \(\sqrt{2g(S-y)}\)

When body is at height y above ground. Potential energy of body of mass m : U = mgy

As per given condition kinetic energy, K = 3U 

\(\frac{1}{2}\) m v² = 3 x mgy

\(\frac{1}{2}\) m 2g(S – y) = 3 x mgy

y = \(\frac{S}{4}\)

v = \(\frac{S}{4}\), \(\sqrt{\frac{3 g S}{2}}\)