A ball of mass m is fired vertically upwards from the surface of the earth with velocity nvₑ , where vₑ is the escape velocity and n < 1. Neglecting air resistance, to what height will the ball rise? (Take radius of the earth as R )

The correct option is: Rn²/(1-n²)

To find the maximum height to which the ball will rise, we can use the law of conservation of mechanical energy. Since air resistance is neglected, the total mechanical energy of the ball-Earth system remains constant throughout the motion.

Let the mass of the Earth be ^M and the radius of the Earth be ^R. The mass of the ball is ^m.
The initial velocity of the ball at the surface of the Earth is given as:
$v=n{v}_e$
where $v_e$ is the escape velocity from the surface of the Earth.

The formula for the escape velocity $v_e$ is:
$v_e=\sqrt{\frac{2GM}{R}}$
Therefore, the square of the initial velocity is:
$v^2=n^2{v}_e^2=n^2\left(\frac{2GM}{R}\right)$

Let $h$ be the maximum height reached by the ball above the surface of the Earth. At this maximum height, the final velocity of the ball becomes zero.
Let us write the conservation of energy equation between the surface of the Earth and the point of maximum height:
$E_{\mathrm{initial}}=E_{\mathrm{final}}$
The initial energy at the surface of the Earth is the sum of its kinetic energy and gravitational potential energy:
$E_{\mathrm{initial}}=\frac{1}{2}m{v}^2-\frac{GMm}{R}$

The final energy at the maximum height $h$ (where distance from the center of Earth is $R+h$ and kinetic energy is zero) is:
$E_{\mathrm{final}}=0-\frac{GMm}{R+h}$

Equating the initial and final mechanical energies:
$\frac{1}{2}m{v}^2-\frac{GMm}{R}=-\frac{GMm}{R+h}$

We can cancel the mass of the ball $m$ from both sides:
$\frac{1}{2}{v}^2-\frac{GM}{R}=-\frac{GM}{R+h}$

Substitute the value of $v^2=n^2\left(\frac{2GM}{R}\right)$ into the equation:
$\frac{1}{2}\left(n^2\frac{2GM}{R}\right)-\frac{GM}{R}=-\frac{GM}{R+h}$

Simplify the kinetic energy term:
$n^2\frac{GM}{R}-\frac{GM}{R}=-\frac{GM}{R+h}$

Divide the entire equation by $\frac{GM}{R}$:
$n^2-1=-\frac{R}{R+h}$

Multiply both sides by -1 to rearrange:
$1-n^2=\frac{R}{R+h}$

Take the reciprocal of both sides:
$\frac{R+h}{R}=\frac{1}{1-n^2}$
$1+\frac{h}{R}=\frac{1}{1-n^2}$

Solve for $\frac{h}{R}$ by subtracting 1 from both sides:
$\frac{h}{R}=\frac{1}{1-n^2}$
-1
$\frac{h}{R}=\frac{1-(1-n^2)}{1-n^2}$
$\frac{h}{R}=\frac{n^2}{1-n^2}$

Multiply both sides by $R$ to find the height $h$:
$h=\frac{R{n}^2}{1-n^2}$

Thus, the maximum height to which the ball will rise is $\frac{R{n}^2}{1-n^2}$.