Q 11-05-204JEE MainJEE Main 2025 (3 Apr, Shift 1)Easy
A particle is released from height $S$ above the surface of the earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively:
Answer: (D) $\dfrac S4,\ \sqrt{\dfrac{3gS}{2}}$
Take potential energy zero at the ground. Total energy $= mgS$. At height $x$: $K = 3U$, so $K + U = 4mgx = mgS \Rightarrow x = \dfrac S4$.
Then $K = \tfrac34mgS = \tfrac12mv^2 \Rightarrow v = \sqrt{\dfrac{3gS}{2}}$.
Solution by Sreeraj P, M.Sc Physics