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(B) Using the principle of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $E

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$\frac{R}{3}$

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(B) Using the principle of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $E_i = K + U = \frac{1}{2}mv^2 - \frac{GMm}{R}$At maximum height $h$: $E_f = 0 - \frac{GMm}{R+h}$Given $v = \frac{v_e}{2} = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}}$.Equating $E_i = E_f$:$\frac{1}{2}m \left(\frac{GM}{2R}\right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$$\frac{3}{4R} = \frac{1}{R+h}$$3(R+h) = 4R$$3R + 3h = 4R$$3h = R$$h = \frac{R}{3}$

$A$ body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body from the surface of the earth is

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