A particle of mass ‘m’ is projected with a velocity v=kVe (k < 1) from the surface of the earth. (Ve=escape velocity) The maximum height above the surface reached by the particle is :

The correct option is:
$\frac{R{k}^2}{1-k^2}$

Step-by-step Explanation:

Let $R$ be the radius of the Earth and $M$ be the mass of the Earth. The escape velocity $V_e$ from the surface of the earth is given by:
$V_e=\sqrt{\frac{2GM}{R}}$

The particle is projected with a velocity $v=k{V}_e$. Therefore, the initial velocity is:
$v=k\sqrt{\frac{2GM}{R}}$

Let $h$ be the maximum height reached by the particle above the surface of the Earth. At this maximum height, the velocity of the particle becomes zero. The distance of this point from the center of the Earth is $r=R+h$.

According to the law of conservation of mechanical energy:
Total Energy at the surface of Earth = Total Energy at the maximum height
$K_{\mathrm{initial}}+U_{\mathrm{initial}}=K_{\mathrm{final}}+U_{\mathrm{final}}$

Substituting the values:
$\frac{1}{2}m{v}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}$

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

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

Simplify the terms:
$k^2\frac{GM}{R}-\frac{GM}{R}=-\frac{GM}{R+h}$

Divide by $GM$ on both sides:
$\frac{k^2}{R}-\frac{1}{R}=-\frac{1}{R+h}$

Rearrange the terms to remove the negative sign on the right:
$\frac{1}{R}\left(1-k^2\right)=\frac{1}{R+h}$

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

Solve for $h$:
$h=\frac{R}{1-k^2}-R$

Take $R$ as a common factor:
$h=R\left(\frac{1}{1-k^2}-1\right)$

Simplify the expression inside the parentheses:
$h=R\left(\frac{1-(1-k^2)}{1-k^2}\right)$

$h=\frac{R{k}^2}{1-k^2}$