Free falling object with air resistance, what's the velocity as a function of time? So, I'm looking into how velocity can be represented as a function of time when the air resistance is also function of velocity. The force of drag is
$$ F_d = C||\vec{v}||\vec{v} $$ 
where $C$ is an arbitrary constant depeding on the object's shape etc.
Obviously, we are assuming a gravitational force of $-mg$ near the surface of the Earth.
What confuses me is that this force itself is not constant to begin with. I have a sense that I can integrate my way out of this, so I'd really appreciate who could tell me where to start.
 A: If it's falling only then you have $F_d=+Cv^2$, where up is the positive direction. You said there is a gravitational force $F_g=-mg$. Write a Newton's 2nd Law equation, set $a=\frac{dv}{dt}$, rearrange, to get dv/g(v) = dt, (I'll let you find $g(v)$) and integrate away.  The $v$ integral is not trivial. Look it up in an integral table, if your teacher will let you do that.
Then you will have to invert the result to solve for the velocity function. That will take some algebraic work.
A: You have a net force 
$$F = F_G - F_D$$
$$ma = -mg + C_0 v^2$$
Dividing both sides by m, letting $C_1 = C_0 / m$,
$$a = -g + C_1 v^2$$
$$\frac{dv}{dt} = -g + C_1 v^2$$
$$\frac{dv}{-g+C_1 v^2} = dt$$
Now to integrate both sides, we need to use the limits $[0,t]$ for $dt$, $[v_0,v]$ for $dv$ (where $v_0$ is our initial velocity.
$$\int_{v_0}^{v} \frac{dv}{-g+C_1 v^2} = \int_{0}^{t} dt$$
Plugging through Wolfram Alpha I get 
$$t = \frac{1}{\sqrt{g C_1}}\cdot[tanh^{-1}(\sqrt{\frac{g}{C_1}}v)-tanh^{-1}(\sqrt{\frac{g}{C_1}}v_0)]$$
which rearranges to
$$v = \sqrt{\frac{C_1}{g}}\cdot tanh[\sqrt{g C_1}t + tanh^{-1}(\sqrt{\frac{g}{C_1}}v_0)]$$
And hence we have velocity as a function of time.
