Question

A missile is launched at an angle of ${60}^o$ to the vertical with a velocity $\sqrt{\frac{0.75G{m}_e}{R_e}}$ from the surface of the earth. Neglecting air resistance, find the maximum height reached by the missile from the surface of the earth. $(m_e=$ mass of the earth, $R_e=$radius of the earth $)$

Answer 1

Kinetic energy $\left(k\right)=\frac{1}{2}m{v}^2$

Conservation of energy $△k=-△v$

$\Rightarrow \frac{1}{2}mv{f}^2-\frac{1}{2}mv{i}^2=-\{\left(\frac{kmM}{R_f}\right)-\left(\frac{-kmM}{R_i}\right)\}$

$\Rightarrow \frac{1}{2}m(v{f}^2-v{i}^2)=-kmM(\frac{1}{R_i}-\frac{1}{R_f})$

$\Rightarrow v{f}^2-v{i}^2=-2km(\frac{1}{R_i}-\frac{1}{R_f})$

$\Rightarrow v{f}^2=v{i}^2-2km(\frac{1}{R_i}-\frac{1}{R_f})$

Initial velocity of vertical component

$\Rightarrow viy=\sqrt{\frac{0.75Gme}{R_e}}\cos 60m/s$

Final velocity of vertical component $vfy=0m/s$

Initial velocity $R_i=R_e$

Final velocity $R_f=?$

$\Rightarrow 0^2=\sqrt{{\left(\frac{0.75Gme}{R_e\cos 60}\right)}^2-2G{m}_e(\frac{1}{R_e}-\frac{1}{R_f})}$

$\Rightarrow 0^2=\left(\frac{0.75G{m}_e}{R_e}\right)\times {\left(\cos 60\right)}^2-2G{m}_e(\frac{1}{R_e}-\frac{1}{R_f})$

$\Rightarrow 0^2=\left(\frac{0.75G{m}_e}{R_e}\right)\times 0.25-\frac{2G{m}_e}{R_e}+\frac{2G{m}_e}{R_f}$

$\Rightarrow \frac{2G{m}_e}{R-f}=1.8125\frac{G{m}_e}{R_e}$

$\Rightarrow \frac{2}{R_f}=\frac{1.8125}{R_e}$

$\Rightarrow R_f=\frac{2R_e}{1.8125}=1.103R_e$

$\therefore$ Final height $=0.103R_e.$