Problem 1. The Basic Equations.

If we let positive velocity be in the downward direction, then the two forces acting on the sky diver are:

force due to gravity, g, is F_g = m g;
force due to air resistance is F_r = - k v;

where m g = is the weight of the sky diver (and chute) and k is the coefficient of friction.

m (dv/dt) = F_g + F_r

that is

dv/dt = g - (k/m) v.

First, we solve the initial value problem with arbitrary k.

Input := 

k = .
g = 32;
m = 155/32;
de1 = v'[t] == g - k/m v[t];
v0 = 0;

Input := 

ans = Flatten[DSolve[{de1, v[0]==v0}, v[t], t]]

Output =

         155         155
{v[t] -> --- - ---------------}
          k     (32 k t)/155
               E             k

Input := 

v[t] /. ans

Output =

155         155
--- - ---------------
 k     (32 k t)/155
      E             k

Input := 

LimVel = %[[1]]

Output =

155
---
 k

Now, we can determine the physical drag coefficients for each of the limiting velocities. This then yields the three different (differential) equations.

Input := 

temp = Flatten[Solve[LimVel==278.8, k]];
k1 = k /. temp

Output =

0.555954

Input := 

temp = Flatten[Solve[LimVel==180.4, k]];
k2 = k /. temp

Output =

0.859202

Input := 

temp = Flatten[Solve[LimVel==21.32, k]];
k3 = k /. temp

Output =

7.27017

Input := 

de1 = v'[t] == g - k1 v[t]/m;
de2 = v'[t] == g - k2 v[t]/m;
de3 = v'[t] == g - k3 v[t]/m;