TITLE: Motion of a falling object with air resistance
QUESTION [7 upvotes]: I was wondering how you would model the velocity of a falling object, taking into account air resistance. Bear in mind I have only studied basic calculus, and have no experience with differential equations.
Considering a general solution for mass $m$ and drag $kv^2$, where $k=\frac{1}{2} \rho AC_D$. Let's say that when $t=0$, $v=0$.
Using Newton's Second Law: $mg - kv^2 = ma$, hence $a = g - \frac{k}{m} v^2$.
From here, I tried to integrate with respect to $t$, which gave $v = gt-\frac{kt}{m}v^2$. This, when solved for v, seems to model the velocity correctly, but I have been led to believe that solutions to differential equations of this kind will involve $e^x$ in some way. Is my solution correct, or have I integrated incorrectly? How do we go from here to a function for v in terms of t?

REPLY [8 votes]: Remembering that $a=\dfrac{dv}{dt}$,
$$m\frac{dv}{dt}=mg-kv^2$$
$$\frac{dv}{mg-kv^2}=\frac{dt}{m}$$
Integrating both sides,
$$\int\frac{dv}{mg-kv^2}=\int\frac{dt}{m}$$
You can find an analytical solution from here.
Here is the full solution on my blog if you are interested:https://philosophicalmath.wordpress.com/2017/10/21/terminal-velocity-derivation/