I'm replying to the OP but I'm quoting Duff Man as what he's said is directly relevant.
The terminal velocity is very difficult to compute since it will be radically different for different shapes (it's a property of how the airflow interacts with the shape that's falling). However, if you can measure or estimate the terminal velocity, you can get an estimate for the parameter "k" (which describes mathematically the "difficult" parts relating to fluid-structure interaction). Once you have k, you can plug it back into your original solution and can estimate how far the skydiver has fallen at any given time.
In short, to answer the question you're asking you will need to know:
a) The terminal velocity for the skydiver
b) How to solve a linear second order ODE
As said the 'k' depends on the shape of the object in question, spheres are different to sky divers, and also depends on the orientation of the object relative to the direction of motion. It gets worse though, it is actually dependent upon the density of the object, the fluid in question and the drag coefficient. The drag coefficient is itself dependent upon the Reynolds number, which is dependent upon relative velocity, object size and fluid viscosity. All of this is covered in depth on Wikipedia, look for the 'drag equation'.
The most mathematically straight forward analysis can be done for a sphere, since it's orientation is irrelevant and the drag coefficient as a function of Reynolds number is known from experiments (I can provide data if so wished). However, a sky diver isn't a sphere, the dynamics will be quite different and little experimental data will be available.
In principle the analysis is the same though, you would integrate the drag equation, including a term for mavity, and that'd give you the answer. You would almost certainly have to do this numerically due to the complex behaviour of k, even for a sphere.
If you make the assumption k is constant then you can do it algebraicly and in fact is something I distinctly remember being covered in M2 A Level mechanics. Whether it is still covered I don't know, I did it circa 2001.
Method :
The system is 1 dimensional so we just write z for height. dz/dt = z' = v and dv/dt = a = z'' (acceleration). Gravitational force is mg (g is negative since it pulls things down). Drag force is kv^2. The initial velocity is 0 and the initial position we take to be height H. Thus using Newton's 2nd law we get
mz'' = kv^2 + mg
Divide by m and let K = k/m
z'' = Kv^2 + g
Using the identity z'' = dv/dt = (dz/dt) (dv/dz) = v (dv/dz) we get
v (dv/dz) = Kv^2 - g
Separate variables so to collect them to get
(v dv)/(Kv^2 + g) = dz
Integrate each side (left as exercise for reader) to get
(1/2K) ln | Kv^2 + g | = z + C
therefore
Kv^2 + g = A exp(2kz)
To compute A use the fact z = H gives v = 0
g = A exp(2kH) so A = g exp(-2kH)
therefore
v^2 = (g/K) ( exp(2k(z-H)) - 1 )
Leave C as C since its algebraicly simpler that way.
This tells you the relationship between height and velocity. Integrating it would give you the relationship between height and time but it's sufficiently unpleasant that the result doesn't instantly spring to mind. You can work out the terminal velocity though, as it's what you'd get for large time, which is equivalent to large (negative) z, which makes the expression reduce to
v^2 -> (g/K) ( - 1 )
Remembering g is negative (so there's no issue with v^2 being proportional to -g) and that K = k/m this means that Kv^2 + g = 0, or equivalently kv^2 + mg = 0, which is precisely what you can deduce from the original equation where kv^2 + mg = 0 tells you z'' = 0, ie no acceleration.
Put in values you can find/know.
