A particle of mass m is projected with a velocity v = KVe ( K < 1)… | MHT CET 2026 (15 May, Shift 2)

Q4MHT CET 2026 · 15 MAY · SHIFT 2Medium

Escape Velocity

A particle of mass $m$ is projected with a velocity $v = KV_e$ ( $K < 1$ ) from the surface of the earth. The maximum height above the surface reached by the particle is ( $V_e$ = escape velocity, $R$ is radius of earth)

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Answer: (c)

Energy conservation: $\frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{R+h}\right)$ .
With $v^2 = K^2v_e^2 = K^2\frac{2GM}{R}$ : $\frac{K^2}{R} = \frac{1}{R} - \frac{1}{R+h}$ .
$\frac{1}{R+h} = \frac{1-K^2}{R} \Rightarrow R+h = \frac{R}{1-K^2}$ .
$h = \frac{RK^2}{1-K^2}$ .
So the answer is (c) $\frac{RK^{2}}{1-K^{2}}$ .

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Escape Velocity question from MHT CET 2026 (15 May, Shift 2)

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