Free-fall with linear drag The standard first-course free-fall with linear drag situation posits a particle falling with a constant acceleration (typically due to approximation of gravity), but with a retarding force that is proportional to the velocity. The equation of motion for the particle is written as
$$m\frac{dv}{dt}=-mg-\lambda v \,\,\,\,\,\,\,\,\textrm{or}\,\,\,\,\,\,\,\,m\frac{d^2y}{dt^2}=-mg-\lambda\frac{dy}{dt}$$
The solution for the velocity as a function of time, $v(t)=\frac{dy}{dt}$, is simply a decreasing exponential plus a constant. On the other hand, the solution for the position as a function of time, $y(t)$, is a decreasing exponential plus a linear term plus a constant, which is a bit more complicated. It's complicated in the sense that the inverse function, $t(y)$, is transcendental (involving the Lambert $W$-function), so solving for the time at which the particle is at a certain height is typically not easy (analytically).
My goal is to somehow change this problem so that, given initial conditions $v(t=0)=0$ and $y(t=0)=h$, I can solve for the velocity at the time/position when/where the particle "hits the floor" (i.e. $v(y=0)=v(t_{\textrm{hit}})$) analytically.
If we multiply the original differential equation by $v^{-1}$, we get
$$m\frac{dv}{v\,dt}=m\frac{dv}{dy}=-\frac{mg}{v}-\lambda$$
This is a nonlinear differential equation with an obvious singularity at $v=0$. I don't know how to solve this, and I don't know how to get a sensible closed-form expression for my desired quantity $v(t_{\textrm{hit}})$. I made a hand-waved plot of what the solutions would look like.

Is it possible to find a nice closed-form solution for $v(t_{\textrm{hit}})$? If so, am I on the right track? Please give suggestions.
 A: Your equation 
$$
m\frac{dv}{dy} = -\frac{mg}{v}- \lambda
$$
is solvable, but it doesn't lead to a closed expression v=v(y). It gives instead y=y(v), which will leave you with a simple looking, but still transcendental equation for v(y=0).  
To solve, separate your variables before formal integration:
$$
\frac{v}{v + \frac{mg}{\lambda}}dv = -\frac{\lambda}{m}dy
$$
Integration obtains then
$$
v - \frac{mg}{\lambda}\ln\left( v + \frac{mg}{\lambda} \right) = -\frac{\lambda}{m}y + C
$$ 
From the initial condition $v=0$ for $y = h$ the integration constant is 
$$
C = \frac{\lambda h}{m} - \frac{mg}{\lambda}\ln\frac{mg}{\lambda}
$$ 
so the final expression becomes 
$$
y(v) = h + \frac{m^2 g}{\lambda^2}\left( \ln\left( \frac{\lambda v}{mg} + 1\right) - \frac{\lambda v}{mg} \right)
$$
Note that since $v<0$ and $\frac{\lambda |v|}{mg} < 1$, the term in $v$ on the rhs is negative as it should. For y=0 we are left with 
$$
\frac{\lambda^2 h}{m^2 g} + \frac{\lambda v_{hit}}{mg} = \ln\left( \frac{\lambda v_{hit}}{mg} + 1\right)
$$
As I said, it's still a transcendental equation, but at least it's only logarithmic.
