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(C) By the law of conservation of energy,the total energy at the surface of the earth equals the total energy at the maximum height $h$.At the surface

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$R / \left( \frac{2gR}{V^2} - 1 \right)$

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(C) By the law of conservation of energy,the total energy at the surface of the earth equals the total energy at the maximum height $h$.At the surface: $E_i = K_i + U_i = \frac{1}{2}MV^2 - \frac{GM_eM}{R}$At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{GM_eM}{R+h}$Equating $E_i = E_f$: $\frac{1}{2}MV^2 - \frac{GM_eM}{R} = - \frac{GM_eM}{R+h}$$\frac{1}{2}V^2 = GM_e \left( \frac{1}{R} - \frac{1}{R+h} \right)$Using $g = \frac{GM_e}{R^2}$,we have $GM_e = gR^2$.$\frac{V^2}{2} = gR^2 \left( \frac{R+h-R}{R(R+h)} \right) = \frac{gR^2h}{R(R+h)} = \frac{gRh}{R+h}$$\frac{V^2}{2gR} = \frac{h}{R+h}$$V^2(R+h) = 2gRh$$V^2R + V^2h = 2gRh$$V^2R = h(2gR - V^2)$$h = \frac{V^2R}{2gR - V^2}$Dividing numerator and denominator by $V^2$: $h = \frac{R}{\frac{2gR}{V^2} - 1}$

$A$ rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V$. Assuming the radius of the earth to be $R$ and negligible air resistance,the maximum height attained by the rocket above the surface of the earth is

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