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## Homework Statement

Suppose a rain drop with mass ##m_0\neq 0## is falling due to gravity with initial velocity ##v_0##, assume ##\frac{dm}{dt}=k=##constant. Solve the differential equation and determine the velocity as ##t\to\infty##

## Homework Equations

##F=\frac{dp}{dt}=\dot{m}v+m\dot{v}##

## The Attempt at a Solution

Since ##F=mg## the D.E. is ##\dot{m}v+m\dot{v}=mg##, substituting in ##\frac{dm}{dt}=k## we find ##kv+m\dot{v}=mg##, since the mass increases with time we have ##m=m_0+kt##. One method I've attempted is essentially using the integrating factor but since the LHS is already a product rule derivative I don't need to multiply it by anything so:

$$\frac{d}{dt}(mv)=mg \\

\int \frac{d}{dt}(mv)dt=\int mg dt=\int mg \frac{dt}{dm}dm=\frac{g}{k}\int_{m_0}^{m} m\, dm\\

v(m_0+kt)+C=\frac{g}{2k}((m_0+kt)^2-m_0^2)=\frac{g}{2k}(2m_0kt+k^2t^2)

$$

at ##t=0##, ##v=v_0\Rightarrow##, ##v_0(m_0+0)+C=0\Longrightarrow C=-v_0m_0##

therefore...

$$v(m_0+kt)-v_0m_0=gm_0t+\frac{gkt^2}{2}\\

v(m_0+kt)=m_0(v_0+gt)+\frac{gkt^2}{2} \\

v=\frac{m_0(v_0+gt)}{m_0+kt}+\frac{gkt^2}{2(m_0+kt)}

$$

Now I find this solution a bit odd since my professor said that it would become asymptotic but I think he's wrong about this? The first term will approach ##\frac{m_0g}{k}## but the second term clearly diverges as ##t\to\infty## and it doesn't make sense that an object would reach a terminal velocity without drag. I suppose you could say it asymptotically goes to infinity but it's increasing linearly...