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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{3gS}{2}

B

\frac{S}{4}, \sqrt{\frac{3gS}{2}}

C

\frac{S}{2}, \sqrt{\frac{3gS}{2}}

D

\frac{S}{4}, \sqrt{\frac{3gS}{2}}

Step-by-Step Solution

Let required height of body is y. When body from rest falls through height (S - y), then under constant acceleration v2=02+2g(S−y)  ⟹  v=2g(S−y)v^2 = 0^2 + 2g(S - y) \implies v = \sqrt{2g(S - y)}. When body is at height y above ground, potential energy U=mgyU = mgy. As per given condition kinetic energy K=3U  ⟹  12m(v)2=3×mg(y)  ⟹  12×m×2g(S−y)=3×mgy  ⟹  S−y=3y  ⟹  y=S4K = 3U \implies \frac{1}{2}m(v)^2 = 3 \times mg(y) \implies \frac{1}{2} \times m \times 2g(S - y) = 3 \times mgy \implies S - y = 3y \implies y = \frac{S}{4}. Then v=2g(S−S4)=3gS2v = \sqrt{2g(S - \frac{S}{4})} = \sqrt{\frac{3gS}{2}}.

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