Ok, I'll do the case of some very fluffy thing where the friction is proportional to the velocity, falling in gravitational acceleration from an initial velocity of zero, reaching terminal velocity V
F. Then after time t the height fallen, h, is given by:
h= V
Ft+(V
F^2/g)(e^(-gt/V
F)-1).
The more common case, like for a ball, where the friction becomes proportional to the square of the velocity after a short period of acceleration is harder to solve. I might have to use a program, or look it up somewhere, if you want that formula too.
Mike W.
So Mathematica didn't download properly, and I had a chance to sleep on it. For the case where the friction is proportional to velocity squared, here's what I get:
time= (V
F/g)*(integral over [1, e^(hg/V
F^2)] of dx/sqrt(x^2-1)
With some help from an integral table:
time=(h/V
F)+(V
F/g)*ln(1+sqrt(1-e^(-2hg/V
F^2))).
I checked the limiting cases (small h and big h) and it looks ok.
Addendum- I'd left out the expression for distance h in terms of time, t, thinking it would be even messier. It actually comes out simple:
h=(V
F^2/g)*ln(cosh(tg/V
F))
Again, limiting cases check.
(published on 04/01/2011)