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

The correct answer is:

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

Step-by-Step Explanation:

Let $R$ be the radius of the Earth, $M$ be the mass of the Earth, and $m$ be the mass of the particle.

The escape velocity ($V_e$) from the surface of the Earth is given by the formula:
$V_e=\sqrt{\frac{2GM}{R}}$
Squaring both sides, we get:
$V_e^2=\frac{2GM}{R}$

The particle is projected from the surface of the Earth with a velocity:
$v=k{V}_e$

At the surface of the Earth, the initial mechanical energy ($E_i$) of the particle is the sum of its kinetic energy and potential energy:
$E_i=K_i+U_i$
$E_i=\frac{1}{2}m{v}^2-\frac{GMm}{R}$
Substituting $v=k{V}_e$ into the equation:
$E_i=\frac{1}{2}m{k}^2{V}_e^2-\frac{GMm}{R}$

Now, substitute $V_e^2=\frac{2GM}{R}$:
$E_i=\frac{1}{2}m{k}^2\left(\frac{2GM}{R}\right)-\frac{GMm}{R}$
$E_i=\frac{GMm{k}^2}{R}-\frac{GMm}{R}$
$E_i=-\frac{GMm}{R}(1-k^2)$

Let $h$ be the maximum height reached by the particle above the surface of the Earth. At this maximum height, the particle's velocity becomes zero, so its final kinetic energy is zero ($K_f=0$).
The final mechanical energy ($E_f$) at distance $R+h$ from the center of the Earth is:
$E_f=K_f+U_f$
$E_f=0-\frac{GMm}{R+h}$

According to the law of conservation of mechanical energy:
$E_i=E_f$
$-\frac{GMm}{R}(1-k^2)=-\frac{GMm}{R+h}$

Simplifying the equation by dividing both sides by $-GMm$:
$\frac{1-k^2}{R}=\frac{1}{R+h}$
Taking the reciprocal of both sides:
$R+h=\frac{R}{1-k^2}$

Solving for $h$:
$h=\frac{R}{1-k^2}-R$
$h=R\left(\frac{1}{1-k^2}-1\right)$
$h=R\left(\frac{1-(1-k^2)}{1-k^2}\right)$
$h=\frac{R{k}^2}{1-k^2}$