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 [

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Solution

Q.
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 [2021]

1 $\frac{S}{4}, \sqrt{\frac{3 g S}{2}}$

2 $\frac{S}{4}, \frac{3 g S}{2}$

3 $\frac{S}{4}, \frac{\sqrt{3 g S}}{2}$

4 $\frac{S}{2}, \frac{\sqrt{3 g S}}{2}$

Ans.
(1)

Let the point be P where K.E. is three times of P.E.

Let the height fall is $x$ and speed at $P$ is $v$ .

$K E_{p} = 3 P E_{p} = 3 m g (S - x)$ ...(i)

Use conservation of energy

$P E$ at top $= P E at P + K E at P$

$m g S = m g (S - x) + K E_{p}$

$K E_{p} = m g S - m g (S - x)$ ...(ii)

From (i) and (ii)

$3 m g (S - x) = m g S - m g (S - x)$

$4 m g (S - x) = m g S$

$4 S - 4 x = S \Rightarrow x = \frac{3 S}{4}$

So, $S - x = S - \frac{3 S}{4} = \frac{S}{4}$

Now, $K E_{p} = \frac{1}{2} m v^{2} = 3 m g (S - x)$

$\frac{v^{2}}{2} = 3 g (S - \frac{3 S}{4}) \Rightarrow \frac{v^{2}}{2} = 3 g \times \frac{S}{4} \Rightarrow v = \sqrt{\frac{3 g S}{2}}$

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