A body is projected up with a velocity equal to $\frac{3}{4} th$ of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth is R) :

\frac{10 R}{9}

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\frac{9 R}{7}

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\frac{9 R}{8}

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\frac{8 R}{5}

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Solution

The correct option is B $\frac{9 R}{7}$
According to the law of conservation of energy
$(T . E)_{at surface} = (T . E)_{at height}$
$\frac{1}{2} m V^{2} + (\frac{- G M m}{R}) = 0 + (\frac{- G M m}{R + h})$
Given $V = \frac{3}{4} V_{e} = \frac{3}{4} \sqrt{\frac{2 G M}{R}}$
$\Rightarrow \frac{1}{2} m \times \frac{9}{16} \times (\frac{2 G M}{R}) + (\frac{- G M m}{R})$
$= (\frac{- G M m}{R + h})$
$\Rightarrow \frac{9 G M m}{16 R} - \frac{G M m}{R} = - \frac{G M m}{R + h}$
$\Rightarrow \frac{- 7 G M m}{16 R} = - \frac{G M m}{R + h}$
$\Rightarrow \frac{7}{16 R} = \frac{1}{R + h}$
$\Rightarrow 7 R + 7 h = 16 R$
$h = \frac{9 R}{7}$

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