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(A) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.Total Energy at surface

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$\frac{R}{8}$

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(A) Using the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.Total Energy at surface = Total Energy at height $h$$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Given $v = \frac{v_e}{3}$,where $v_e = \sqrt{\frac{2GM}{R}}$ is the escape velocity.Substituting $v^2 = \frac{v_e^2}{9} = \frac{2GM}{9R}$:$\frac{1}{2}m\left(\frac{2GM}{9R}\right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{GMm}{9R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$Dividing by $GMm$:$\frac{1}{9R} - \frac{1}{R} = - \frac{1}{R+h}$$\frac{1-9}{9R} = - \frac{1}{R+h}$$\frac{-8}{9R} = - \frac{1}{R+h}$$\frac{8}{9R} = \frac{1}{R+h}$$8(R+h) = 9R$$8R + 8h = 9R$$8h = R$$h = \frac{R}{8}$

$A$ body is projected vertically upwards from the Earth's surface of radius $R$ with a velocity equal to $\frac{1}{3}$ of the escape velocity. The maximum height reached by the body is

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