Q138 / 150 · MHT CET 2018 · Physics

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)

Correct answer
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
$\frac{u^{2} R}{g R - u^{2}}$

Solution

\frac{- G M m}{R} + \frac{1}{2} m u^{2} = 0 + (- \frac{G M m}{R + h})

\frac{G M}{R + h} = \frac{G M}{R} - \frac{u^{2}}{2}

\frac{G M}{(R + h)} = \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}}

Hence, the correct answer is option: 'A'.

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