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The escape velocity of a body on the earth’s surface is $V_{e}$ . A body is thrown vertically up with a speed of $k V_{e} (w h e r e k < 1)$ . The maximum height reached by the body above the earth is (‘R’ is the radius of the earth)

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$R (\frac{1 - k^{2}}{k})$

$\frac{R}{k^{2}}$

$R^{2} \frac{k^{2}}{(1 - k^{2})}$

$R (\frac{k^{2}}{1 - k^{2}})$

answer is A.

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Detailed Solution

$v_{C} = \sqrt{\frac{2 G M}{R}}$

according to 2CE,

$\Rightarrow - \frac{G M m}{R} + \frac{1}{2} m v^{2} = \frac{- G M m}{R + h} + 0 \Rightarrow h = \frac{R k^{2}}{1 - k^{2}}$

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