A particle is released from height $S$ above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

<p><strong>Using the equation of motion for velocity</strong> at height $ x $: </p> <p>$ V^2 = 0 + 2g(S - x) $</p> <p>Simplifies to:</p> <p>$ V^2 = 2g(S - x) $</p></p> <p><p><strong>Potential Energy</strong> at height $ x $: </p> <p>$ \text{Potential Energy} = mgx $</p></p> <p><p><strong>Relation between kinetic and potential energy</strong>: </p> <p>Given that the kinetic energy is three times the potential energy,</p> <p>$ mgx = 3 \times \frac{1}{2} mV^2 $</p> <p>Simplifying gives:</p> <p>$ gx = \frac{3}{2} \times 2g(S - x) $</p> <p>$ gx = 3g(S - x) $</p> <p>$ 4x = S $</p> <p>Solving for $ x $:</p> <p>$ x = \frac{S}{4} $</p></p> <p><p><strong>Finding the speed $ V $ at this instant</strong>:</p> <p>$ V = \sqrt{2g \left( S - \frac{S}{4} \right)} = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}} $</p></p> <p>Thus, the height from the surface of the Earth is $ \frac{S}{4} $ and the speed of the particle at that height is $ \sqrt{\frac{3gS}{2}} $.</p>