Question

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.

• A. \( \frac{S}{2} \), \( \sqrt{\frac{3gS}{2}} \)
• B. \( \frac{S}{2} \), \( \frac{3gS}{2} \)
• C. \( \frac{S}{4} \), \( \sqrt{\frac{3gS}{2}} \)
• D. \( \frac{S}{4} \), \( \frac{3gS}{2} \)

Hint:

Use the conservation of energy principle to relate the potential and kinetic energies at different heights. Remember that the potential energy is proportional to the height above the surface.

Answer 1

The Correct Option is C

\( V^2 = 0 + 2g(S-x) \) \( V^2 = 2g(S-x) \) 

At B, Potential energy = mgx Kinetic energy 

= \( \frac{1}{2} mv^2 \) \( \frac{1}{2} mv^2 = 3mgx \) 

\( gx = \frac{1}{6} v^2 = \frac{1}{6} 2g(S-x) \) \( 4x = S \) 

\( x = \frac{S}{4} \) \( V = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}} \)