Ok, lets separate the vertical and horizontal movements of the bomb.
In the vertical direction, the bomb is left to fall freely from 15 km, with Earth’s gravity accelerating it towards the ground. We can therefore calculate how long will it take to reach the ground using the formula:
s = s0 + v0·t + 1/2 · a · (t*t)
Where s is the position of the bomb, s0 is the initial position, v0 is the initial speed (in the vertical direction, that is, zero), a is the acceleration (this is g = 9.8 m/s*s) and t is time. So, in our case:
0 = 15000 + 0·t + 1/2 · (-9.8) · (t*t)
Solving the second degree equation you get t = 55.328 s, wich is the time elapsed from the moment the bomb is dropped to the moment it reaches the ground.
Now, after that time, and considering that the bomb is flying in the horizontal direction at a constant speed of 1593 km/h (ie 442,5 m/s), the bomb have traveled
s = s0 + v0 · t
Wich, in our case is
s = 0 + 442.5 · 55.328 = 24482,7876 m.
As you can see, the weight of the bomb is not important at all.
NOTE: We have ignored other forces that may be actuating on the bomb, specially air friction. In any case, to take air friction into acount will make the bomb travel even LESS than 24.5 km so, by any means can it reach 39 km.
You will need to:
- Fly faster, or
- Release the bomb closer to the target, or
- Go bombing to another planet, where gravity is less intense.