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.
Show Hint
- \( \frac{S}{2} \), \( \sqrt{\frac{3gS}{2}} \)
- \( \frac{S}{2} \), \( \frac{3gS}{2} \)
- \( \frac{S}{4} \), \( \sqrt{\frac{3gS}{2}} \)
- \( \frac{S}{4} \), \( \frac{3gS}{2} \)
The Correct Option is C
Solution and Explanation
\( V^2 = 0 + 2g(S-x) \) \( V^2 = 2g(S-x) \)
At B, Potential energy = mgx Kinetic energy
= \( \frac{1}{2} mv^2 \) \( \frac{1}{2} mv^2 = 3mgx \)
\( gx = \frac{1}{6} v^2 = \frac{1}{6} 2g(S-x) \) \( 4x = S \)
\( x = \frac{S}{4} \) \( V = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}} \)
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