Q 11-05-009NEETNEET 2021Top questionMedium
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: (A) $\dfrac{S}{4},\ \sqrt{\dfrac{3gS}{2}}$
Concept. Total energy $mgS$ is shared as $K + U$.
Step 1. Height. $K = 3U \Rightarrow 4mgh = mgS \Rightarrow h = \dfrac S4$.
Step 2. Speed. $\dfrac12mv^2 = \dfrac34mgS \Rightarrow v = \sqrt{\dfrac{3gS}{2}}$.
Answer. $\dfrac S4,\ \sqrt{\dfrac{3gS}{2}}$ (Option A).
Solution by Sreeraj P, M.Sc Physics