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Physics Question on work, energy and power

A particle is released from a height SS . 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

S4,3gS2\frac{S}{4},\frac{3gS}{2}

B

S4,3gS2\frac{S}{4},\frac{\sqrt{3gS}}{2}

C

S2,3gS2\frac{S}{2},\frac{3gS}{2}

D

S4,3gS2\frac{S}{4},\sqrt{\frac{3gS}{2}}

Answer

S4,3gS2\frac{S}{4},\sqrt{\frac{3gS}{2}}

Explanation

Solution

We can realise the situation as shown. Let at point C distance xx from highest point A, the particles kinetic energy is three times its potential energy.
Velocity at C, v2=0+2gx{{v}^{2}}=0+2gx or v2=2gx{{v}^{2}}=2gx ...(i)
Potential energy at C =mg(S−x)=mg(S-x) ...(ii)
At point C, Kinetic energy
=3×=3\times potential energy ie,
12m×2gx=3×mg(S−x)\frac{1}{2}m\times 2gx=3\times mg(S-x) or x=3S−3xx=3S-3x
or 4x=3S4x=3S
or S=43xS=\frac{4}{3}x
or x=34Sx=\frac{3}{4}S
Therefore, from E (i) v2=2g×34S{{v}^{2}}=2g\times \frac{3}{4}S
Or v2=32gS{{v}^{2}}=\frac{3}{2}gS
or V=32gSV=\sqrt{\frac{3}{2}gS}
Height of the particle from the ground
=S−x=S-x
=S−34S=S4=S-\frac{3}{4} S=\frac{S}{4}