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(B) The velocity of projection is $v = \frac{3}{4} v_e$,where $v_e = \sqrt{2gR}$ is the escape velocity.Using the principle of conservation of energy:

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$\frac{9 R}{7}$

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(B) The velocity of projection is $v = \frac{3}{4} v_e$,where $v_e = \sqrt{2gR}$ is the escape velocity.Using the principle of conservation of energy:$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Substituting $v = \frac{3}{4} \sqrt{\frac{2GM}{R}}$:$\frac{1}{2}m(\frac{9}{16} \cdot \frac{2GM}{R}) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{9GMm}{16R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$$-\frac{7GMm}{16R} = - \frac{GMm}{R+h}$$\frac{7}{16R} = \frac{1}{R+h}$$16R = 7(R+h) = 7R + 7h$$7h = 9R$$h = \frac{9}{7}R$

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

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