Ok, I tried scouring the forum for a reasonable answer to my question... (SPOILER OBV) Brownie points to the person who explains it the best:
The blast radius of the nuke is 6 miles. When Batman decides to tow the bomb away, I believe there is about 1:43 left on the clock.
For argument's sake, let's assume that:
1. Batman flew and towed the bomb in a straight line -- no detours or evasive maneuvers -- toward and over the water.
2. The Bat reached maximum velocity instantaneously, i.e. no acceleration, and maintained that velocity the entire way.
3. The distance of the flight was exactly 6 miles from the departure point, i.e. the bomb detonated at the absolute minimum safe distance from Gotham.
4. The Bat flew at the max velocity, in a straight line, for 103 seconds. Batman didn't waste any time attaching the tow, walking to and from the bomb, macking on Catwoman, entering the Bat, and/or starting the Bat.
Under the above assumptions, that means the Bat had to have been traveling 209.7 MPH for 103 seconds in a straight line. Throw in factors like acceleration, maneuvering, attachment, lift-off, etc., it had to have traveled at a greater maximum velocity than 209 MPH.
Can someone please explain this sequence in a way that makes it physically possible.... in other words, how can everything be reconciled logically?
It did not look like the Bat could travel any faster than about 50-60 MPH laterally; even a 120 MPH max velocity with zero acceleration and zero detour would mean that Bats needed 3+ minutes to do it.