The initial horizontal velocity of the areoplane is equal to the speed of the aeroplane. Therefore, it can be said that when the bomb is released, the bomb too moves forward.
Horizontal velocity, vx = 600 km/h = \(600times frac{5}{18}\) m.s-1 = 166.67 m.s-1
Vertical velocity, vy = 0
Height, h = 1960m
Let time taken to fall = t
\(h=v_{y}t+frac{1}{2}gt^{2}\)1960 = 0 + (½) x (9.8) x (t2)
1960 = 4.9t2
t2 = (1960/4.9) = 400
t = \(sqrt{400}\) = 20s
Therefore, the horizontal distance travelled between the points A and B is:
AB = vxx t = 166.67 x 20 = 3333.4m
