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(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $

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

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(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $E_i = \frac{1}{2}mV^2 - \frac{GMm}{R}$At maximum height $h$: $E_f = 0 - \frac{GMm}{R+h}$Equating $E_i = E_f$: $\frac{1}{2}mV^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$Using $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this:$\frac{V^2}{2} = gR^2 (\frac{1}{R} - \frac{1}{R+h})$$\frac{V^2}{2} = gR^2 (\frac{R+h-R}{R(R+h)}) = \frac{gRh}{R+h}$$\frac{V^2}{2gR} = \frac{h}{R+h}$Taking the reciprocal: $\frac{2gR}{V^2} = \frac{R+h}{h} = \frac{R}{h} + 1$$\frac{R}{h} = \frac{2gR}{V^2} - 1$$h = \frac{R}{(\frac{2gR}{V^2} - 1)}$

$A$ rocket is fired vertically upwards from the surface of the Earth with a velocity $V$. If $R$ is the radius of the Earth,what is the maximum height attained by the rocket?

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