A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :

Solution

Physics sir · 3 steps

The bomb leaves with the plane's own horizontal speed, so relative to the plane it only falls straight down.

1. At release. When the bomb is let go, it is already moving horizontally with the plane's speed u. Nothing pushes it sideways after that, so its horizontal velocity stays u (ignoring air resistance).
2. Relative velocity. The plane also keeps moving at u horizontally. Subtract the plane's velocity from the bomb's velocity: the horizontal parts cancel, and only the vertical part gt is left.

   v⃗_(bomb, plane) = (u î - gt ĵ) - u î = -gt ĵ

3. So. Seen from the plane, the bomb starts from rest and accelerates downward at g, always staying directly below the plane. Its path is a straight line vertically down, option (c).

The idea to keep. Always subtract the observer's velocity first. If both bodies share the same horizontal velocity, that motion vanishes for the observer and only the vertical fall remains.

Answer from the NTA final answer key. Solution by PrepWiser.