Q. A particle is released from a height$\ S$. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively
A) $\frac{S}{4},\frac{3gS}{2}$
B) $\frac{S}{4},\frac{\sqrt{3gS}}{2}\ $
C) $\frac{S}{2},\frac{\sqrt{3gS}}{2}$
D) $\frac{S}{4}.\frac{\sqrt{3gS}}{2}$
Explanation
(d) Velocity at $B$ when dropped from $A$ where $AC = S$
$v^{2} = 0 + 2g(S - x)$
Or $v^{2} = 2g(S - x)$ …(i)
Potential energy at $B = mgx$ …(ii)
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$\because$ Kinetic energy $= 3 \times$ potential energy
$\therefore\frac{1}{2}\ m \times 2g(S - x) = 3 \times mgx$
$\Rightarrow S - x = 3x$ or $S = 4x$ or $x = S/4$
From (i),
$v^{2} = 2g(S - x) = 2g\left( S - \frac{S}{4} \right) = \frac{2g \times 3S}{4} = \frac{3gS}{2}$
$\Rightarrow v = \sqrt{\frac{3gS}{2}}\ \ \therefore x = \frac{S}{4}$ and $v = \sqrt{\frac{3gS}{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.