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Physics · PYQ NEET UG 2021
हिन्दीA particle of mass 'm' is projected with a velocity \( v = kv_e \) (\( k < 1 \)) from the surface of the earth. (\( v_e \) = escape velocity). The maximum height above the surface reached by the particle is:
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Correct answer · Option C
\( \frac{Rk^2}{1 - k^2} \)
Explanation
By conservation of mechanical energy: \( -\frac{GMm}{R} + \frac{1}{2}m(kv_e)^2 = -\frac{GMm}{R+h} \). Since \( v_e = \sqrt{\frac{2GM}{R}} \), substituting yields \( h = \frac{Rk^2}{1-k^2} \).
