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  72km/hr. The car is at a distance of 500m from the gun when the guns fired at an angle of  ${45}^0$ with the horizontal. The  distance of the car from the gun when the shell hits the car is _____m. $(g=9.8m/s^2)$

The shell hits the car after a time T sec
At a distance ‘R’ from the gun.
 

$R=\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin {90}^0}{g}=\frac{u^2}{g}$
Time of flight,  $T=\frac{2u\sin {45}^0}{g}=\frac{\sqrt{2}u}{g}$
According to problem 
$$\begin{gathered} R=500+VT V=\frac{72\times 5}{18}=20 \\ \frac{u^2}{g}=500+20\left(\frac{\sqrt{2}u}{g}\right) \\ {u}^2-20\sqrt{2}u-500g=0 \\ {u}^2-20\sqrt{2}u-500\times 9.8=0 \\ \therefore u=10[\sqrt{2}\pm \sqrt{51}] \\ Taking +ve sign \\ u=10[\sqrt{2}+\sqrt{51}]=85.56m/s \\ Dis\tan ce, R=\frac{u^2}{g}=\frac{{\left(85.56\right)}^2}{9.8} =747m \end{gathered}$$