Question

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 1

Let, the 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)

When the body is at height y above the ground. Potential energy of body of mass m,

U = mgy

As per given condition kinetic energy, K = 3U

⇒ \(\frac 12 m(v)^2 = 3 \times mg (y)\)   .....(2)

Using equations (1) and (2),

⇒ S − y = 3y

∴ y = \(\frac S4\)

From (1),

\(v = \sqrt{2 \times g(S - \frac S4)}{}\)

\(= \sqrt{\frac {3gS}2}\)

Answer 2

Option : (4)

U + KE = E

4U = E = mgS

4mgh = mgS

h = \(\frac{S}{4}\)