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Home » 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 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
Physics | Physics
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

(1)\,\,\frac{S}{4},\frac{3gS}{2}

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

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

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

Solution: Answer (4)

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)}1/2 …(1)

When body is at height y above ground. The potential energy of body of mass m

U = mgy

As per given condition kinetic energy, K = 3U

\frac{1}{2}m{{v}^{2}}=3mgy

\frac{1}{2}m\times 2g(S-y)=3\times mgy

S-y=3y

y=\frac{S}{4}

\therefore v=\sqrt{2g\left( S-\frac{S}{4} \right)}=\sqrt{\frac{3gS}{2}}

i

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