First Order ODE – A skydiver weighing 180 lb falls vertically downward ...from an altitude of 5000ft and opens the parachute after 10s of free fall. Assume that the force of air resistance, which is directed opposite velocity, is of magnitude 0.75|v| when the parachute is closed and is of magnitude 12|v| when the parachute is open, where velocity is measured in ft/s.
a) Find the speed of the skydiver when the parachute opens.
b) Find the distance fallen before the parachute opens
For the first part of have set up the answer as follows:
F=ma <=>
180$\frac{dv}{dt}$ = mg-0.75v where v(0) = 0 I got the solution
v(t) = $7680 e^{(0.00417 t)}-7680$. When substituting t = 10s, I got the answer 327 m/s, whereas the answer given is 176.7 m/s. Can someone explain how to get the actual answer? And can someone also explain how to do the second part of the question? Thanks!
 A: First, your equation is correct, $m\dot v=mg-bv$. The solution of which with $v(0)=0$ is
$$
v(t)-\frac{mg}b= e^{-b/m·t}\left(v_0-\frac{mg}b\right)\iff v(t)=\frac{mg}b\left(1-e^{-b/m·t}\right)
\approx g·t-\frac{gb}{2m}·t^2,
$$
the latter approximation for small $t$, corresponding in first order to the free fall without friction. In the approximation, the velocity after $10s$ is $98.1{\rm [m/s]}=321.8{\rm [ft/s]}$, all more exact values must be close to this.
With $m=180 {\rm[lb]}=81.6[\rm kg]$, $g=9.81 [\rm m/s^2]$ and $b=0.75\rm [kg·m/s^2·s/ft] = 0.2286\rm [kg/s]$ the solution with friction gives
$$
v(t)=3 501.7 {\rm[m/s]}·\left(1-e^{-0.00280{\rm[1/s]}·t}\right)
$$
where I get $v(10{\rm[s]})=96.69 \rm{[m/s]}=317.2\rm{[ft/s]}$.
Now it could be that the friction force constant is actually measured in $\rm [lb·ft/s^2·s/ft]=[lb/s]$, however, the task description is vague on the units used.
The proposed solution $327$ could be correct if measured in $\rm[ft/s]$ and with a gravitation constant of $10{\rm[m/s^2]}=32.81{\rm[ft/s^2]}$.
