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Q.

A projectile is fired from the surface of earth of radius $R$ with a velocity $k v_{e}$ $($ where $v_{e}$ , is the escape velocity from surface of earth and $k < 1)$ . Neglecting air resistance, the maximum height of rise from the centre of earth is

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a

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

b

$k^{2} R$

c

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

d

$k R$

answer is C.

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Detailed Solution

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{1}{2} m {(k v_{e})}^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{r}$

$∴ \frac{1}{2} k^{2} (2 g R) - \frac{G M}{R} = - \frac{G M}{r}$

or $\frac{G M}{r} (1 - k^{2}) = \frac{G M}{r} (as g R = \frac{G M}{R})$

So, $r = \frac{R}{1 - k^{2}}$

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