Question

a bomber is flying with a constant speed of 150 m/s at the height of 78.4 m the pilot has to drop a bomb at the enemy target at what horizontal distance from the target should he release the bomb so that it hits the target

Answer 1

D=u√(2H/g)
=150×√(156.8/9.8)
=150×4
=600m
Option C is correct

Answer 2

This is a sum of vectors. The horizontal velocity of the bomb will be unaffected by the vertical acceleration and velocity of the bomb when it is dropped. Let's first consider the vertical velocity portion.

h=ut+½gt²

The plane (with the bomb) flies at height h=78.4m
initial vertical velocity of the bomb u=0 (before it is dropped, it remains at a condition of VERTICAL rest, but moves with constant velocity along the horizontal axis)
so,
h=0+½gt²
=> 78.4=½×9.8×t²
=> t²=16
=> t=4

Now, since horizontal vector is unaffected by gravitational force, the bomb, even after it is dropped, will fall with same HORIZONTAL velocity 150m/s as it had when it was contained inside the plane. So it will fall will horizontal velocity 150m/s. It takes 4 secs to fall, and within that time, it will cover a distance 150m/s×4 secs = 600m.
So while falling, it will advance 600m due to its unaffected horizontal momentum. Thus it must be dropped 600m in advance. (c)