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(A) Let $h$ be the maximum height reached by the object. According to the law of conservation of energy,the total energy at the surface of the earth i

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$R/3$

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(A) Let $h$ be the maximum height reached by the object. According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$.Total energy at the surface: $E_i = KE_i + PE_i = \frac{1}{2} m v_p^2 - \frac{GMm}{R}$,where $v_p$ is the projection velocity.Given that the projection velocity is half the escape velocity: $v_p = \frac{1}{2} v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.Thus,$KE_i = \frac{1}{2} m (\frac{1}{2} \sqrt{\frac{2GM}{R}})^2 = \frac{1}{2} m (\frac{1}{4} \cdot \frac{2GM}{R}) = \frac{GMm}{4R}$.Total energy at the surface: $E_i = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}$.At maximum height $h$,the velocity is zero,so $KE_f = 0$. The potential energy is $PE_f = -\frac{GMm}{R+h}$.Equating initial and final energies: $-\frac{3GMm}{4R} = -\frac{GMm}{R+h}$.$\frac{3}{4R} = \frac{1}{R+h} \implies 3(R+h) = 4R \implies 3R + 3h = 4R \implies 3h = R \implies h = R/3$.

$A$ certain object is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half the escape velocity. The maximum height attained by the object will be

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