Terminal velocity of falling sphere with drag and buoyancy

[Celso]

Homework Statement

Show that the terminal speed of a falling spherical object is given by $v_{t} = [ (mg/c_{2})+(c_{1}/2c_{2})^2]^{1/2} - (c_{1}/2c_{2})$ when both the linear and the quadratic terms in the drag force are taken into account.

Relevant Equations

$m\ddot x = c_{2}v^2 + c_{1}v - mg$

To write $v$ as a function of time, I wrote the equation $m\frac{dv}{dt} = c_{2}v^2 + c_{1}v - mg \implies \frac{mdv}{c_{2}v^2 + c_{1}v - mg} = dt$
To solve this, I thought about partial fractions, but several factors of $-c_{1} \pm \sqrt {c_{1}^2 +4c_{2}*mg}$ would appear and they don't show up in the resolution of the problem statement

 

[BvU]

Terminal speed is a constant $\Rightarrow$ no need to solve for $v(t)$. Just looking at ${dv\over dt}=0$ is sufficient.

 

[RPinPA]

Your expression and the given solution are closely related.

Follow the approach suggested by @BvU at #2 and it pops right out, both your expression and the actual solution.