A body attains a height equal to the radius of the earth. The velocity of the body with which it was projected is

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$\sqrt{(\frac{GM}{R})}$

$\sqrt{(\frac{2 GM}{R})}$

$\sqrt{(\frac{5 GM}{4 R})}$

$\sqrt{(\frac{3 GM}{R})}$

answer is A.

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Detailed Solution

Applying the law of conservation of energy,, we have
$\frac{1}{2} {mu}^{2} - \frac{GMm}{R} = \frac{1}{2} m (0)^{2} - \frac{GMm}{R + R}$
or $u^{2} = \frac{2 GM}{R} - \frac{2 GM}{2 R} = \frac{GM}{R}$
$∴ u = \sqrt{(GM / R)}$

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