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

To find the maximum height attained by the rocket, we can apply the law of conservation of mechanical energy. Since air resistance is negligible, the total mechanical energy of the rocket remains constant during its flight.

Let:
- $M$ be the mass of the rocket,
- $R$ be the radius of the Earth,
- $M_E$ be the mass of the Earth,
- $V$ be the initial launch speed of the rocket from the surface of the Earth,
- $h$ be the maximum height attained above the surface of the Earth,
- $g$ be the acceleration due to gravity at the Earth's surface, where $g=\frac{G{M}_E}{R^2}$ or $G{M}_E=g{R}^2$.

At the surface of the Earth (initial state):
The distance from the center of the Earth is $R$.
The kinetic energy is:
$K_i=\frac{1}{2}M{V}^2$
The gravitational potential energy is:
$U_i=-\frac{G{M}_{E}M}{R}=-MgR$

At the maximum height $h$ (final state):
The distance from the center of the Earth is $R+h$.
At maximum height, the velocity of the rocket is zero, so the kinetic energy is:
$K_f=0$
The gravitational potential energy is:
$U_f=-\frac{G{M}_{E}M}{R+h}=-\frac{Mg{R}^2}{R+h}$

Using the conservation of energy:
$K_i+U_i=K_f+U_f$
Substitute the values into the equation:
$\frac{1}{2}M{V}^2-MgR=0-\frac{Mg{R}^2}{R+h}$

We can divide the entire equation by the mass of the rocket $M$:
$\frac{1}{2}$

Rearranging the equation to solve for $R+h$:
$\frac{g{R}^2}{R+h}=gR-\frac{1}{2}$
Taking the reciprocal of both sides:
$\frac{R+h}{g{R}^2}=\frac{1}{gR-\frac{1}{2}}$
Multiplying both sides by $g{R}^2$:
$R+h=\frac{g{R}^2}{gR-\frac{1}{2}}$

To simplify, subtract $R$ from both sides:
$h=\frac{g{R}^2}{gR-\frac{1}{2}}-R$
$h=\frac{g{R}^2-R\left(gR-\frac{1}{2}\right)}{gR-\frac{1}{2}}$
$h=\frac{g{R}^2-g{R}^2+\frac{1}{2}}{gR-\frac{1}{2}}$
$h=\frac{\frac{1}{2}}{gR-\frac{1}{2}}$

Multiply the numerator and denominator by $\frac{2}{V^2}$ to format it like the options:
$h=\frac{R}{\frac{2gR}{V^2}-1}$

Thus, the maximum height attained by the rocket above the surface of the earth is:
$R/\left(\left(2gR/V^2\right)-1\right)$