Sorry I was gone so long. When I went to put together a general solution, my brain complained "You haven't asked me for that kind of information since you retired! I don't remember where I put it. I need a road map to get that deep. Why don't you ask that computer you keep sending me information from?" It took a while to get my brain reoriented, and now I find that FiveFinger knows so much math that my technique may not be the best thing to suggest.

So let me comment on the accuracy of the expressions for the drag force first. I did find an article online which described an actual measurement of the drag force on a small particle. It was rather technical, but it showed a graph that looked pretty accurately F = c v

^{2} to me. The curve was flat enough that F = bv would have been a decent approximation for smaller values of v. Usually F = c v

^{2} is justified in textbooks by a rather rough argument, and drag is complicated enough that you wouldn't expect a single, simple formula to work in all cases. F = bv is mostly justified as a low-v approximation, and it is used solely because you can write explicit solutions for it using only elementary functions like the exponential.

Now, what I was going to do. A general solution, with 4 undetermined constants, for the problem when F = mg - b dy/dt, and using g>0 and downward positive, is y = B

_{1} e

^{-At} + B

_{2} + B

_{3}. Since

F

_{total} = ma = m d

^{2}/dy

^{2} and F

_{total} = mg - b dy/dt

y must satisfy the equation

m d

^{2}y/dt

^{2} + b dy/dt = mg

Differential equation theory tells us that y must have the form

y = B

_{1} e

^{-At} + B

_{2}t + B

_{3}and FiveFingers deserves more explanation than that, which I can do later if desired. Happily there are 4 conditions available to determine the 4 constants:

1. y(0) = y

_{0} = the starting point

2. v(0) = dy(0)/dt = v

_{0} = the starting velocity

3,4. Substitute into the differential equation. The resulting equation must be true at all times or we haven't gained anything. There are two functions of time in the equation, constant#1 e

^{-At} and constant#2 t. Their sum is 0, and that can be true for all time only if constant#1 and constant#2 are both equal to 0 [for instance, try t=0 and t=infinity].

For maximum satisfaction, I would suggest evaluating this form and checking numerically that it works, then using the more general form to do the numerical solution again. You might not care about the second numerical solution after doing the first one, since you will then know that you can do it if you ever need to.

Edit: I get

"An expert is a person who has already made all possible mistakes." -- paraphrase of a statement by Niels Bohr

Seen on a bumper sticker: "My other vehicle is a Krebs cycle".