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

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TestHub · Accepted Answer

We can realize the situation as shown. Let at point $C$ distance $x$ from highest point $A$ , the particle's kinetic energy is three times its potential energy.

Velocity $C,$

$v^{2} = 0 + 2 g x$

$O r v^{2} = 2 g x$ .....(i)

Potential energy at , $C = m g (S - x) \dots . . (i i)$

At Point C,

Kinetic energy $= 3 \times$ potential energy

ie, $\frac{1}{2} m \times 2 g x = 3 \times m g (S - x)$

or $x = 3 S - 3 x$

or $4 x = 3 S$

or $S = \frac{4}{3} x$

or $x = \frac{3}{4} S$

Therefore, from Eq.(i)

$v^{2} = 2 g \times \frac{3}{4} S$

Or $v^{2} = \frac{3}{2} g S o r v = \sqrt{\frac{3}{2} g S}$

Height of the particle from the ground

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

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