Wobble, the bomb has a combined movement. 2 axis. The vertical one is moving the bomb away from the buff. The horizontal one keeps the bomb steady with the buff. Let's say the buff flies at 250 kts. . Everything inside the buff is moving at 250 kts. Inside the plane, lift counters gravity, so vertical speed of everything inside equals zero (i'm not counting falling pencils from navigator's desk

). Now you open doors, release the bomb. At this time, the bomb is still moving at 250 kts. on the X axis. Only thing that changes is that now, unattached to the plane, the bomb generates practically 0 lift force, so besides the 250 kts. in the X-axis, it's building speed on the vertical axis. Given the 9'8 m/s^2 gravitational constant, each second the bomb is falling, it gains 9'8 m/s in vertical speed. The now freefalling bomb is moving on the path determined by the resultant of 2 vectors. One is horizontal and paralell to the Earth, the X-axis speed that, if we forget about drag, is constant at 250 kts. The other one is vertical, perpendicular to the Earth, and determines a uniformly accelerated movement. So, in the second zero, the bomb is moving 250 kts. on the horizontal, and 0 kts. on the vertical (flat trajectory). The first second X-speed is 250 Kts., and Y-speed is 9'8 m/s. The second nr.2 X-speed is 250kts. and the Y-speed is 19'6 m/s. and so on. So the resultant for the freefalling bomb is a parabolic movement. This does not mean that the bomb will fall behing because it travels farther as, with regards to the plane, the bomb is not moving in the X-Axis (equal speeds) while it goes down.
So, if we do not put drag into the equation, relative X-position of the falling bomb and the buff, given the buff does not changes its course, speed or altitude, remains constant. If we introduce the drag, we alter the X-axis component of the freefall movement. Instead of a straight, constant speed movement, we have a straight, uniformly deccelerated one. The magnitude of the decceleration relates to the drag, and this aspect relates to the shape & mass of the bomb. Given the bomb is a fusiform object (close to the ideal aerodinamic shape, which is the one the drop of water adopts in a freefall), and very massive (thus neglecting the flotability - lift issues), I would say drag forces can be safely neglected without affecting hardly the "sim" side of this game.
So, as a conclusion, and correct me if I'm wrong, the bomb would fall exactly under the buff, if we negate drag's influence, and slightly behind the buff, if drag is modelled.
Cheers,
Pepe