Height and Velocity of a Falling Particle – Rankers Physics
Topic: Work Energy and Power
Subtopic: Potential Energy & Equilibrium

Height and Velocity of a Falling Particle

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

Solution:

Total mechanical energy is \(E = mgS\). At height \(h\), \(KE = 3 PE⇒ E = KE + PE = 4 PE ⇒ mgS = 4mgh ⇒ h = S/4\). Also, \(\frac{1}{2}mv^2 = 3mg(S/4) ⇒ v = \sqrt{\frac{3gS}{2}}\).

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