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(A) 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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(A) 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: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$Since $E_i = E_f$,we have $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$Given $v = \frac{v_e}{2} = \frac{1}{2}\sqrt{\frac{2GM}{R}}$,so $v^2 = \frac{GM}{2R}$.Substituting $v^2$ into the energy equation:$\frac{1}{2}m(\frac{GM}{2R}) - \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} \Rightarrow 3(R+h) = 4R \Rightarrow 3R + 3h = 4R \Rightarrow 3h = R \Rightarrow h = \frac{R}{3}$

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

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