Q. 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:
A) $\frac{S}{4}, \frac{\sqrt{3gS}}{2}$
B) $\frac{S}{2}, \frac{\sqrt{3gS}}{2}$
C) $\frac{S}{4}, \sqrt{\frac{3gS}{2}}$
D) $\frac{S}{4}, \frac{3gS}{2}$
Explanation
[Solution] mgS = KE + PE = 4 mgh, so h = S/4. KE = mg(3S/4), so $v = \sqrt{3gS/2}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.