Question: 24. 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 :
(1) \frac{S}{4}, \frac{3gS}{2}
(2) \frac{S}{4}, \frac{\sqrt{3gS}}{2}
(3) \frac{S}{2}, \frac{\sqrt{3gS}}{2}
(4) \frac{S}{4}, \sqrt{\frac{3gS}{2}}
Answer: Option (4)
Explanation:
Let the particle be released from rest from a height S above the surface of the Earth.
Total mechanical energy of the particle at the top is purely potential energy and is given by E = mgS.
Suppose at a certain height h from the surface, the kinetic energy is three times the potential energy.
So, \text{K.E.} = 3 \times \text{P.E.}.
The potential energy at height h is mgh and the kinetic energy is mg(S-h), since the loss in potential energy is converted into kinetic energy.
Thus, mg(S-h) = 3mgh.
Canceling mg from both sides, we get S - h = 3h.
This gives S = 4h or h = \frac{S}{4}.
The kinetic energy at this height is \frac{1}{2}mv^{2} = mg(S-h).
Substituting h = \frac{S}{4},
\frac{1}{2}mv^{2} = mg\left(S - \frac{S}{4}\right) = mg\frac{3S}{4}.
Simplifying, v^{2} = \frac{3gS}{2}.
Hence, the speed of the particle is v = \sqrt{\frac{3gS}{2}}.
Therefore, the height from the surface and the speed of the particle are \frac{S}{4}
and \sqrt{\frac{3gS}{2}} respectively, corresponding to Option (4).