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
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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)