A gun kept on a straight horizontal road is used to shoot at a car travelling on the same road away from the gun at a uniform speed of 102 ms-1. The car is at a distance of 150 m from the gun when it is fired at an angle of 45° to the horizontal. With what speed should the shell be projected so that it hits the car? Take g =10 ms-2.

Let at t = 0, O and A are the positions of the gun and the car. Let us say that at time T, the shell and the car reach B simultaneously so that the shell hits the car when it is at a distance OB from the gun. Let u be the speed of projection of the shell. Then, initial horizontal component of velocity of the shell =u cos ⁡45∘=u2and initial vertical component =u sin ⁡45∘=u/2. The car takes this time to cover the distance AB while the shell covers the distance OB in this time. Now OB = OA + AB = 150 m + AB. Distance AB is given by AB=102×2×ug=20u/gAnd       OB=u2×2ug=u2/g∴        u2g=150+20ugor       u2−20u−1500=0 ∵g=10 ms−2The positive root of this quadratic equation is u = 50 ms-1.