Problem source

A particle of unit mass is projected vertically upwards in a medium whose resistance is $k$ times the square of the velocity of the particle. If the initial velocity is $u$, prove that the velocity $v$ after rising through a distance $s$ satisfies 
\begin{equation*}
v^{2}=u^{2}\e^{-2ks}+\frac{g}{k}(\e^{-2ks}-1). \tag{*}
\end{equation*}
Find an expression for the maximum height of the particle above  he point of projection. Does equation $(*)$ still hold on the downward path?
Justify your answer.

Solution source

\begin{align*}
\text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g - kv^2 \\
\Rightarrow && \int \frac{v}{g+kv^2} \d v &= \int -1 \d s \\
\Rightarrow && \frac{1}{2k}\ln(g+kv^2) &= -s + C \\
s =0, v = u: && \frac{1}{2k} \ln(g+ku^2) &= C \\
\Rightarrow && s &= \frac{1}{2k} \ln \frac{g+ku^2}{g+kv^2} \\
\Rightarrow && e^{-2ks} &= \frac{g+kv^2}{g+ku^2} \\
\Rightarrow && v^2 &= u^2e^{-2ks} + \frac{g}{k}(e^{-2ks}-1)
\end{align*}

The maximum height will be when $v = 0$, ie  $\displaystyle s = \frac{1}{2k}\ln\left(1 + \frac{k}{g}u^2 \right)$.

On the downward path the resistance will be going upwards, ie
\begin{align*}
\text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g + kv^2
\end{align*}
but our solution is solving a different differential equation, therefore unless $k=0$ the equation will be different.