Out of curiosity and boredom, I'm going to do a little bit of math. Now, I know this doesn't actually mean anything in a D&D world, and especially in Rich's comic strip, but like I said... I'm bored and curious.
First, let's assume that physics in general works the same as the real world...
There are 16 frames of Roy falling until he hits the ground. At 3 seconds (1 half-turn) per frame (to account for the actions he's taking), that makes for 48 seconds of falling... We'll round it up to 50 for simplicity's sake, and to account for any time-slop at the beginning and end of the fall.
In general, a person in free-fall has a terminal of approximately 54 m/s.
t = v/g = 54/9.8 = 5.5 seconds.
d = ½*g*t² = ½*9.8*5.5² = 296.45 m = 973 ft.
So, Roy reaches terminal velocity and falls nearly a thousand feet in the first round.
Too bad he's still got another 7 rounds (42 seconds) to fall at terminal velocity...
d = v*t = 54*42 = 2,268 m = 7,441 ft.
Using paper-napkin physics, that would have been about an 8,400 ft. fall Roy just took.
Ouch.
