JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2025
- Easy
- MCQ
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.
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Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{S}}{4}, \sqrt{\frac{3 \mathrm{gS}}{2}}\)
Step-by-step Solution
Detailed explanation
Potential energy at height \( h \): \( U = mgh \)
Initial mechanical energy: \( E = mgS \)
Kinetic energy at height \( h \): \( K = mg(S - h) \)
Given condition: \( K = 3U \)
\( mg(S - h) = 3mgh \)
\( S - h = 3h \Rightarrow S = 4h \)
\( h = \frac{S}{4} \)
Speed at height \( h \): \( \frac{1}{2}mv^2 = mg\left(S - \frac{S}{4}\right) \)
\( \frac{1}{2}mv^2 = mg\frac{3S}{4} \)
\( v = \sqrt{\frac{3gS}{2}} \)
Initial mechanical energy: \( E = mgS \)
Kinetic energy at height \( h \): \( K = mg(S - h) \)
Given condition: \( K = 3U \)
\( mg(S - h) = 3mgh \)
\( S - h = 3h \Rightarrow S = 4h \)
\( h = \frac{S}{4} \)
Speed at height \( h \): \( \frac{1}{2}mv^2 = mg\left(S - \frac{S}{4}\right) \)
\( \frac{1}{2}mv^2 = mg\frac{3S}{4} \)
\( v = \sqrt{\frac{3gS}{2}} \)
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