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 :
Answer & explanation
Correct answer: option 3
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 = \(\frac{S}{4}\), \(\sqrt{\frac{3 g S}{2}}\)