How to recover free fall equation from equations of motion with linear drag

Question

For a particle of mass $m$, initial velocity $v = 0$, and initial height $h$, integrate the equation of motion to find $y(t)$, assuming that the particle experiences a frictional force that is linearly proportional to its velocity, namely $F=−αv$. Demonstrate explicitly that as $\alpha\rightarrow0$ you recover the free fall equation.

I started with $m\frac{dv}{dt} = -mg - \alpha v$, and I arrived at the following equation: $$y(t) = h - \frac{m^2g}{\alpha^2}e^{-\frac{\alpha}{m}t} - \frac{mg}{\alpha}t$$ It is easy to find that the velocity approaches terminal velocity, and the unbounded position approaches $-\infty$. However, I can't see a readily apparent way to recover the free fall equation from the position equation. Simply setting $\alpha$ to $0$ doesn't help because I don't have correct units, and I arrive at $y(t) = h - 0 - \infty$. I don't see any simple algebra tricks, and a Taylor expansion doesn't seem to help either. Could someone help me find the error in my approach please?

Answer 1

The trouble comes from a mistake in solving the equation $$m\frac{d^2y}{dt^2} = -mg - \alpha \frac{dy}{dt}\quad\text{with conditions}\quad y(0)=h\text{ and }y'(0)=0.$$ You wrote : $$y(t) = h - \frac{m^2g}{\alpha^2}e^{-\frac{\alpha}{m}t} - \frac{mg}{\alpha}t$$ In fact, a term is missing. The correct solution is : $$y(t) = h - \frac{m^2g}{\alpha^2}e^{-\frac{\alpha}{m}t} - \frac{mg}{\alpha}t +\frac{m^2g}{a^2}$$ Expending to series with respect to $\alpha$ leads to : $$y=h-\frac12 gt^2+\left(\frac{g}{6m}\right)t^3\alpha+O\left(t^4\alpha^2\right)$$ In case of $\alpha=0$, one recovers $y=h-\frac12 gt^2$.

Answer 2

$$ v'(t)=-\frac{\alpha}{m}v(t)-g $$ Using as integrating factor $Q=e^{\alpha t/m}$ we have $$ \frac{(Qv)'}{Q}=-g $$ and integrating $$ Qv=-\frac{mgQ}{\alpha}+\frac{c}{Q} $$ and then $$ v(t)=-\frac{mg}{\alpha}+\left(v(0)+\frac{mg}{\alpha}\right)e^{-\alpha t/m} $$ from which we have $$ y(t)=y(0)-\frac{mg}{\alpha}t+\frac{m}{\alpha}\left(v(0)+\frac{mg}{\alpha}\right)\left(1-e^{-\alpha t/m}\right) $$ With $y(0)=h,\,v(0)=0$ we have $$ y(t)=h-\frac{mg}{\alpha}t+\frac{m^2g}{\alpha^2}\left(1-e^{-\alpha t/m}\right) $$ Using $e^{-\alpha t/m}\sim 1-\frac{\alpha t}{m}+\frac{\alpha^2 t^2}{2m^2}+o(\alpha^3)$ as $\alpha\to 0$ we have \begin{align} y(t)&\sim h-\frac{mg}{\alpha}t+\frac{m^2g}{\alpha^2}\left(1-1+\frac{\alpha t}{m}-\frac{\alpha^2 t^2}{2m^2}+o(\alpha^3)\right)\\ &\sim h-\frac{g}{2}t^2 \end{align}