A projectile is projected with velocity kv_e in vertically upward direction from the ground into the space. ( v_e is escape velocity and k < 1) . If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

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

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

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

$\frac{R}{k + 1}$

answer is C.

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Detailed Solution

Kinetic energy = Potential energy
$\frac{1}{2} m {({kv}_{e})}^{2} = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow \frac{1}{2} {mk}^{2} 2 gR = \frac{mgh}{1 + \frac{h}{R}}$
$\Rightarrow h = \frac{{Rk}^{2}}{1 - k^{2}}$
Height of Projectile from the earth's surface = h
Height from the centre $r = R + h = R + \frac{{Rk}^{2}}{1 - k^{2}}$
By solving $r = \frac{R}{1 - k^{2}}$ .