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


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.

enter image description here

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.

  • $\begingroup$ The linear term has nothing to do with the problem, at all. It's just a transformation into a completely equivalent inertial system, which is physically irrelevant. $\endgroup$
    – CuriousOne
    Sep 15, 2015 at 0:16
  • $\begingroup$ I can't see how that would simplify the problem. Maybe you could expand on it? The linear term has coefficient $-mg/\lambda$, so from what I understood, you are suggesting to make the transformation $y'=y+(mg/\lambda)t$, $t'=t$. In that inertial system, there would be no linear term, but now we would be solving for $y'(t_{\textrm{hit}})$, and that isn't necessarily equal to 0 as it is in the original inertial frame. $\endgroup$ Sep 15, 2015 at 0:33
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    $\begingroup$ @ArturodonJuan: you seem to want to needlessly want to complicate things. $m\frac{d^2y}{dt^2}=-g-\lambda\frac{dy}{dt}$ is easily solved. Form there any information you want should be deducible. $\endgroup$
    – Gert
    Sep 15, 2015 at 0:55
  • $\begingroup$ @Gert I know that that equation is easily solvable. I even stated what the solutions will behave like. What I'm saying is that to solve for the velocity of the particle when it has "hit the ground" (i.e. $y=0$) doesn't seem to be an easy task. I can't see how to get a closed-form solution for that particular quantity. $\endgroup$ Sep 15, 2015 at 1:04
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    $\begingroup$ @ArturodonJuan - If by "closed form solution" you mean "expressible as a finite combination of elementary functions", the answer is you can't. You already showed that the inverse function is non-elementary. End of story. $\endgroup$ Sep 15, 2015 at 1:10

1 Answer 1


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.


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