The velocity of an aeroplane is 600 km/h which is flying horizontally and at a height of 1960m. A bomb is released from the aeroplane when it vertically reaches point A. At point B, the bomb is struck. What is the distance between AB?

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

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