Question

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 $20$ m/s. The car is at a distance of $160$m from the gun, when the gun is fired at an angle of ${45}^o$ with the horizontal. Find the distance of the car from the gun when the shell hits it and the speed of projection of the shell from the gun.

Answer 1

Let the speed of the shell fired from the gun be 'v' $\frac{m}{s}$

Angle at which the shell is fired = 45${}^{\circ }$

Time of flight of the shell = $\frac{2vsin\Theta }{g}$

= $\frac{2vsin\left({45}^{\circ }\right))}{g}$

= $\frac{v\sqrt{2}}{g}$

Distance covered by car during this time = 20*t

= 20*$\frac{v\sqrt{2}}{g}$

Range of shell = $\frac{v^{2}sin\left(2\Theta \right)}{g}$

=$\frac{v^2}{g}$

For the shell to hit the car,

160 + 20t = $\frac{v^2}{g}$

160 + 20*$\frac{v\sqrt{2}}{g}$ = $\frac{v^2}{g}$

Take g = 10 $\frac{m}{s^2}$

We get v = 40$\sqrt{2}$ $\frac{m}{s}$

Hence, distance of the car from the gun when the shell hits it = 160 + 20t

= 320 m

And, speed of projection from the gun = 40$\sqrt{2}$ $\frac{m}{s}$