Two paper screens A and B (standing vertically) are horizontally separated by a distance of 100 m. A bullet pierces A and then B. The hole in B is 10 cm below the hole in A. If the bullet is travelling horizontally at the time of hitting the screen A, calculate the velocity of the bullet when it hits the screen A. Take g = 9.8 m.s⁻²

Let the horizontal component of the speed of the bullet when it passes through A is v m•s−1. Now it's vertical component is 0m•s−1. It falls down a height of 0.1m (10 cm). Now time taken to travel this vertical distance, Vertical distance = (-0.1 m) Initial vertical component (u) = 0 m•s−1 S = u*t - 1/2(g*t²) Now S = -1/2 (g*t²) -0.1 = -4.9 * t² Cancelling the negative sign t = 1/7 seconds Now horizontally time taken by the bullet to travel the horizontal distance is same Distance = Speed*Time 100 = v * (1/7)v = 700 m•s−1