Answer
$v = (\dfrac{m g}{k})^2$ (Terminal Velocity)
Work Step by Step
$ma=m -k g \sqrt v $
Re-write as: $a= g -\dfrac{k\sqrt v}{m}$
$\dfrac{dv}{dt}=g -\dfrac{k\sqrt v}{m}$
When $\dfrac{dv}{dt}=0 $, then, we have:
$g -\dfrac{k\sqrt v}{m}=0$
or, $v = (\dfrac{m g}{k})^2$
This is known as the Terminal Velocity.
