Question

A body is thrown from the surface of the earth with velocity $u \, m \, s^{- 1}$ . The maximum height in meter above the surface of the earth up to which it will reach is ( $R= \, $ radius of the earth, $g=$ acceleration due to gravity)

• A. $\frac{u^{2} R}{2 g R - u^{2}}$
• B. $\frac{2 u^{2} R}{g R - u^{2}}$
• C. $\frac{u^{2} R^{2}}{2 g R^{2} - u^{2}}$
• D. $\frac{u^{2} R}{g R - u^{2}}$

Answer 1

Answer: (A)

$\frac{- G M m}{R} + \frac{1}{2} m u^{2} = 0 + \left(\right. - \frac{G M m}{R + h} \left.\right)$
$\frac{G M}{R + h} = \frac{G M}{R} - \frac{u^{2}}{2}$
$\frac{G M}{\left(\right. R + h \left.\right)} = \frac{2 GM - R u^{2}}{2 R}$
$\frac{R + h}{G M} = \frac{2 R}{2 G M - R u^{2}}$
$h = \frac{2 G M R}{2 G M - R u^{2}} - R$
$= \frac{2 G M R - 2 G M R + R^{2} u^{2}}{2 G M - R u^{2}}$
$= \frac{R^{2} u^{2}}{2 G M - R u^{2}} = \frac{R u^{2}}{2 g R - u^{2}}$