Hi MarcoP,
I'll give it a shot, just don't rely on my answer :))
Find the velocity of a falling body at t=5 seconds
weight of the body=50kg
Air resistance is proportional to the velocity of the body.
weight of the body=50kg =m
\(\begin{align}\\air\;\; resistance\;\; up \;&\;\alpha \;\; velocity \;\;down\\ mass*accel\;&\;\alpha \;\; velocity \;down \\ m\frac{dv}{dt}&=k*\frac{dx}{dt} \\ \frac{dv}{dt}&=\frac{k}{m}*\frac{dx}{dt} \end{align} \)
Height = 100m
Initial velocity = 5m/sec
Limiting velocity = 65m/sec
Let downwards be positive velocity
\(Initially\\ \;\;t=0\;\;\;\;\ x=100 \;\; \;\; \dot x =5 \;\; \;\;\ddot x=g-\frac{k}{50}*5=g-0.1k\)
Ongoing
\(\begin{align}\\\frac{dv}{dt}&=g-\frac{k}{50}*v\\ \frac{dv}{dt}&=\frac{50g-kv}{50}\\ \frac{dt}{dv}&=\frac{50}{50g-kv}\\ t&=\frac{50ln(50g-kv+c_1)}{-k}\\~\\ &When\;\;t=0 \;\;v=5\;\;find\;\; c_1\\~\\ 0&=\frac{50ln(50g-5k+c_1)}{-k}\\ 0&=50ln(50g-5k+c_1)\\ 1&=50g-5k+c_1\\ c_1&=1-50g+5k\\ \frac{-kt}{50}&=ln(50g-kv+1-50g+5k)\\ \frac{-kt}{50}&=ln(1-kv+5k)\\ e^{\frac{-kt}{50}}&=1-kv+5k\\ &As \;\;t\rightarrow\infty \quad v\rightarrow 65\\~\\ 0&=1-65k+5k\\ 0&=1-60k\\ 60k&=1\\ k&=\frac{1}{60}\\ so\\ e^{\frac{-t}{60*50}}&=1-\frac{v}{60}+\frac{5}{60}\\ e^{\frac{-t}{3000}}&=\frac{13}{12}-\frac{v}{60}\\ &When\;\; t=5\\ e^{\frac{-5}{3000}}&=\frac{13}{12}-\frac{v}{60}\\ e^{\frac{-1}{600}}&=\frac{65}{60}-\frac{v}{60}\\ 60e^{\frac{-1}{600}}&=65-v\\ v&=65-60e^{\frac{-1}{600}}\\ v&=5.1 \end{align}\\ \)
Which is wrong :(
Alan can you see my error with any ease ? 
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