Question

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. $\frac{S}{2},\frac{3gS}{2}$
• D. $\frac{S}{4},\sqrt{\frac{3gS}{2}}$
• E. $\frac{S}{3},\sqrt{\frac{3gS}{2}}$

Answer 1

Answer: (D)

We can realise the situation as shown. Let at point C distance $x$ from highest point A, the particles kinetic energy is three times its potential energy.
Velocity at C, $v^2=0+2gx$ or $v^2=2gx$ ...(i)
Potential energy at C $=mg(S-x)$ ...(ii)
At point C, Kinetic energy
$=3\times$ potential energy ie,
$\frac{1}{2}m\times 2gx=3\times mg(S-x)$ or $x=3S-3x$
or $4x=3S$
or $S=\frac{4}{3}x$
or $x=\frac{3}{4}S$
Therefore, from Eq. (i) $v^2=2g\times \frac{3}{4}S$
Or $v^2=\frac{3}{2}gS$
or $V=\sqrt{\frac{3}{2}gS}$
Height of the particle from the ground
$=S-x$
$=S-\frac{3}{4}S=\frac{S}{4}$