We have a 500km/h horizontally moving aeroplane travelling in the x direction. When the plane is directly over its objective, a bomb is dropped from the plane. The distance between the target and the plane are given as 1000 metres. We know that if a bomb is dropped from a moving plane, it will have a horizontal velocity. As a result, the bomb will miss the target and land at a distance of ‘x’ from it.

We can say that the initial velocity of the bomb in the vertical direction will be zero.

vy(initial)=uy=0

By the third kinematic equation, we have s = ut + 1/2at2

Here displacement, ‘s’ in y direction is 1000m, initial velocity is zero and acceleration

‘a’ is the acceleration due to gravity, g=10m/s2.

Therefore, sy = uyt+1/2gt2

1000 = 0 + 1/2 ×10 × t2

t = 10√2sec

Now we have time of flight t=10√2 sec

We have the horizontal velocity, v=500km/hv = 500 × 5/18v =138.889m/s

Therefore the distance from the target to the bomb ‘x’ will be x=v×tx=138.889×10√2

x=1964m

x=1.964km

**Hence the bomb missed the target by 1.96km.**