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(C) The initial speed of the rocket is $v_i = 0.5 v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.At maximum height $h$,the final velocity $v_f = 0$.Applying

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$\frac{R}{3}$

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(C) The initial speed of the rocket is $v_i = 0.5 v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.At maximum height $h$,the final velocity $v_f = 0$.Applying the law of conservation of mechanical energy:$K_i + U_i = K_f + U_f$$\frac{1}{2} m v_i^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Substituting $v_i^2 = \frac{1}{4} \left( \frac{2GM}{R} \right) = \frac{GM}{2R}$:$\frac{1}{2} m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\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}$

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ($R$ - Radius of the earth).

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