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) $\dfrac{S}{4},\ \sqrt{\dfrac{3gS}{2}}$
Total energy $= mgS$. When $K = 3U$, $K + U = 4U = mgS$, so
$$U = mgh = \frac{mgS}{4} \;\Rightarrow\; h = \frac{S}{4}$$
$$K = \frac{1}{2}mv^2 = \frac{3mgS}{4} \;\Rightarrow\; v = \sqrt{\frac{3gS}{2}}$$
Solution by Sreeraj P, M.Sc Physics