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Q.

A body is projected up with a velocity equal to 3/4^th of the escape velocity from the surface of the earth. The height it reaches is (radius of the earth = R)

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a

$\frac{10 R}{4}$

b

$\frac{9 R}{8}$

c

$\frac{9 R}{7}$

d

$\frac{10 R}{3}$

answer is B.

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Detailed Solution

$V = \frac{3}{4} V_{e} \Rightarrow V_{e} = \sqrt{2 gR}$
max. ht. attained by the body is
$H = \frac{V^{2} R}{2 gR - V^{2}} = \frac{9 R}{7}$

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