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Q.

A body is projected horizontally from the surface of the earth $(radius = R)$ with a velocity equal to n times the escape velocity. Neglect rotational effects of the earth. The maximum height attained by the body from the earth’s surface is $R/2$ . Then n must be

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a

0.6

b

(\sqrt{3}) / 2

c

\sqrt{0.4}

d

1 / 2

answer is A.

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Detailed Solution

According to L.C.E

$\frac{- GMm}{R} + \frac{1}{2} {mn}^{2} v_{e}^{2} = \frac{- GMm}{R + h}$ $[h = \frac{R}{2}]$

$\frac{- GMm}{R} + \frac{1}{2} {mn}^{2} \frac{2GM}{R} = \frac{- 2 GMm}{3R}$

$- 1 + n^{2} = \frac{- 2}{3}$

$n^{2} = 1 - \frac{2}{3} = \frac{1}{3}$

$n = \frac{1}{\sqrt{3}} = 0.6$

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