$A$ rocket is fired vertically from the surface of Mars with a speed of $2 \; km/s$. If $20 \%$ of its initial kinetic energy is lost due to Martian atmospheric resistance,how far will the rocket go from the surface of Mars before returning to it? (Mass of Mars $= 6.4 \times 10^{23} \; kg$,radius of Mars $= 3395 \; km$,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$)

(495 KM) Initial velocity of the rocket,$v = 2 \; km/s = 2 \times 10^3 \; m/s$.
Mass of Mars,$M = 6.4 \times 10^{23} \; kg$.
Radius of Mars,$R = 3395 \; km = 3.395 \times 10^6 \; m$.
Gravitational constant,$G = 6.67 \times 10^{-11} \; N m^2 kg^{-2}$.
Initial kinetic energy $K_i = \frac{1}{2} m v^2$.
Initial potential energy $U_i = -\frac{GMm}{R}$.
Since $20 \%$ of initial kinetic energy is lost,the effective kinetic energy available is $0.8 K_i = 0.8 \times \frac{1}{2} m v^2 = 0.4 m v^2$.
Total initial energy $E_i = 0.4 m v^2 - \frac{GMm}{R}$.
At maximum height $h$,the final velocity is zero,so final energy $E_f = -\frac{GMm}{R+h}$.
By conservation of energy,$E_i = E_f$:
$0.4 m v^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$.
Dividing by $m$ and rearranging: $0.4 v^2 = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMh}{R(R+h)}$.
Solving for $h$: $\frac{R+h}{h} = \frac{GM}{0.4 v^2 R} \implies \frac{R}{h} = \frac{GM}{0.4 v^2 R} - 1$.
$h = \frac{0.4 R^2 v^2}{GM - 0.4 v^2 R}$.
Substituting values: $h = \frac{0.4 \times (3.395 \times 10^6)^2 \times (2 \times 10^3)^2}{(6.67 \times 10^{-11} \times 6.4 \times 10^{23}) - (0.4 \times (2 \times 10^3)^2 \times 3.395 \times 10^6)}$.
$h = \frac{18.442 \times 10^{18}}{42.688 \times 10^{12} - 5.432 \times 10^{12}} = \frac{18.442}{37.256} \times 10^6 \approx 495 \; km$.