A gun kept on a straight horizontal road is used to hit a car travelling along the same road away from the gun with a uniform speed of 72 km h^-1. The car is at a distance of 500m from the gun, when the gun is fired at an angle of 45° with the horizontal. Find the speed of projection of the shell from gun. $\text{ (in }\left.\mathrm{m}/{\mathrm{s}}^{-1}\right)\mathrm{ }(\sqrt{2}=1.414,\sqrt{51}=7.141)$

 

Consider that at the instant, the shell is fired from the gun at point O, the car is at point P The shell will move along parabolic path for T its time of flight and will then strike the ground at point Q at a distance R, equal to the horizontal range. Therefore, the shell will hit the car, if in the time T (time of flight), the car moves the distance PQ = v T, where v is velocity of the car. Since the car is initially at a distance of 500 m from the gun, it follows that
R = 500 + vT .....(1)
(a) Let u be the speed of projection of shell.

Now, $\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}},\text{ Here }\mathrm{\theta }={45}^{\circ }$
Also,  $\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\times {45}^{\circ }}{9.8}=\frac{{\mathrm{u}}^2\sin {90}^0}{9.8}=\frac{{\mathrm{u}}^2}{9.8}$
Further, velocity of the car, $\mathrm{v}=72\mathrm{km} {\mathrm{h}}^{-1}$
$=72\times \left(1000\mathrm{m}\right)\times (60\times 60\mathrm{s}{)}^{-1}=20{\mathrm{ms}}^{-1}$
Substituting for R, T and v in equation (1), we have 
$\frac{{\mathrm{u}}^2}{9.8}=500+20\times \frac{\sqrt{2}\mathrm{u}}{9.8}\text{ or }{\mathrm{u}}^2-\sqrt{2}\mathrm{u}-500+9.8=0$
on solving for u we have $\mathrm{u}=10(\sqrt{2}+\sqrt{51})$
$\mathrm{u}=10(1.414+7.141)10\times 8.555=85.55{\mathrm{ms}}^{-1}$