the horizontal distance travelled by a bullet fired at 45 degree is 360 m. if it is fired from a lorry moving in the direction of bullet with a uniform velocity 18 KM/HR and with the same elevation, what is the horizontal distance travelled by the bullet

 

Given that when θ = 45°, horizontal range, R=360 m.

Let u be the velocity with which the bullet is fired and v be the velocity of the lorry.

 

Given that the lorry is moving with a uniform velocity, v= 18 km/ hr =(18×5/18)m/s = 5 m/s

In horizontal direction, as the motion is uniform the acceleration is zero. 

So we can write the displacement in the horizontal direction as R = u_x × T

Here u_x = u cos θ+ v

        u_y = u sin θ

Hence Horizontal range R = (u cosθ+v)× T ---------(1)

 

The horizontal distance travelled by the bullet = 402.9 m