Explanation

(c) If $m$ is the mass of racket, $M$ that of earth and $R$ is the radius of earth, then gravitational potential energy of racket near the surface of earth
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{1} = \frac{GMm}{R}$
Gravitational potential energy of racket at a height$\ h$ from earth’s surface
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{2} = - \frac{GMm}{(R + h)}$
Increase in gravitational potential energy of racket
$\ \ \ \ \mathrm{\Delta}U = U_{2} - U_{1} = - \frac{GMm}{R + h} + \frac{GMm}{R}$
$or\ \ \ \ \mathrm{\Delta}U = \frac{GMmh}{(R + h)R}$
If $v$ is the escape velocity of racket, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = \frac{1}{2}{mv}^{2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}{mv}^{2} = \frac{GMmh}{(R + h)R}$
$\Longrightarrow \ \ {mv}^{2}R^{2} + {mv}^{2}Rh = 2GMm\ h$
$\Longrightarrow \ \ \ \ \ \ v^{2}R^{2} = \left( 2GM - v^{2}R \right)h$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = \frac{v^{2}R^{2}}{2GM - v^{2}R}$