Problem source

Two identical particles of unit mass move under gravity in a medium for which the magnitude of the retarding force on a particle is $k$ times its speed. The first particle is allowed to fall from rest at a point $A$ whilst, at the same time, the second is projected upwards with speed $u$ from a point $B$ a positive distance $d$ vertically above $A$. Find their distance apart after a time $t$ and show that this distance tends to the value 
\[
d+\frac{u}{k}
\]
as $t\rightarrow\infty.$

Solution source

Both particles have equations of motion, $\ddot{x} = -g-k\dot{x}$, so we can note that the distance between them has the equation of motion:

$\ddot{x} = -k \ddot{x} \Rightarrow x = Ae^{-kt} + B$

\begin{align*}
&& x(0) &= d \\
\Rightarrow && A+B &= d \\
&& x'(0) &= u \\
\Rightarrow && -kA &= u \\
\Rightarrow && A &= -\frac{u}{k} \\
\Rightarrow && B &= d+\frac{u}{k} \\
\Rightarrow && x(t) &= -\frac{u}{k}e^{-kt} + d + \frac{u}{k} \to d + \frac{u}{k}
\end{align*}

as required.