Particle released from height S – Rankers Physics

Principal of Conservation of Energy: Practice Problem & Solution

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

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Principal of Conservation of Energy. Understanding the underlying formula is key to arriving at the correct answer below:

Let the initial height be \(S\). At height \(h\), \(KE = 3PE = 3mgh\). By conservation of energy, \(mgS = mgh + 3mgh = 4mgh\), so \(h = S/4\). Also, \(KE = \frac{1}{2}mv^2 = 3mgh\), substituting \(h\) gives \(v^2 = 6g(S/4) = \frac{3gS}{2}\), so \(v = \sqrt{\frac{3gS}{2}}\).

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