Problem source

\textit{In this question, take the value of $g$ to be $10\ \mathrm{ms}^{-2}.$}
A body of mass $m$ kg is dropped vertically into a deep pool of liquid. Once in the liquid, it is subject to gravity, an upward buoyancy force of $\frac{6}{5}$ times its weight, and a resistive force of $2mv^{2}\mathrm{N}$ opposite to its direction of travel when it is travelling at speed $v$ $\mathrm{ms}^{-1}.$ Show that the body stops sinking less than $\frac{1}{4}\pi$ seconds after it enters the pool. 
Suppose now that the body enters the liquid with speed $1\ \mathrm{ms}^{-1}.$ Show that the body descends to a depth of $\frac{1}{4}\ln2$ metres and that it returns to the surface with speed $\frac{1}{\sqrt{2}}\ \mathrm{ms}^{-1},$ at a time 
\[
\frac{\pi}{8}+\frac{1}{4}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)
\]
seconds after entering the pool.

Solution source

While descending, the body experiences the force $-\frac15mg - 2mv^2$. 

\begin{align*}
    \text{N2:} && m \dot{v} &= -\frac15 mg - 2mv^2 \\
    \Rightarrow && \frac{\dot{v}}{\frac15g + 2v^2} &= -1 \\
    \Rightarrow && \frac{1}{2}\tan^{-1} v_1 - \frac{1}{2}\tan^{-1} {v_0} &= -T 
\end{align*}

We care about when $v_1 = 0$, ie $\displaystyle T = \frac{1}{2}\tan^{-1} {v_0} < \frac12 \frac{\pi}2 = \frac{\pi}4$ seconds.

If the body enters at speed  $1\ \mathrm{ms}^{-1}.$ then for the first part of it's journey it will experience forces  $-\frac15mg - 2mv^2$ and so:

\begin{align*}
    \text{N2:} && m v \frac{\d v}{\d x} &= -\frac15 mg - 2mv^2 \\
    \Rightarrow && \int \frac{v}{2(1 + v^2)} \d v &= \int -1 \d x \\
    \Rightarrow && \frac14 \ln (1 + v^2) &= -x 
\end{align*}

Therefore the depth is $\frac14 \ln 2$ metres. 

When the body is rising, it experiences forces of: $\frac15mg - 2mv^2$ and so:

\begin{align*}
    \text{N2:} && m v \frac{\d v}{\d x} &= \frac15mg - 2mv^2 \\
    \Rightarrow && \int \frac{v}{2(1 - v^2)} \d v &= \int -1 \d x \\
    \Rightarrow && -\frac14 \ln (1 - v^2) &= \frac14 \ln 2 \\
    \Rightarrow && 1-v^2 &= \frac12 \\
    \Rightarrow && v &= \frac{1}{\sqrt{2}} \ \mathrm{ms}^{-1}
\end{align*}

This will take

\begin{align*}
    \text{N2:} && m \dot{v} &= \frac15mg - 2mv^2 \\
    \Rightarrow && \frac{\dot{v}}{2(1-v^2)} &= -1 \\
    \Rightarrow && \dot{v} \frac{1}{4}\l \frac{1}{1 - v} + \frac{1}{1+v} \r &= -1 \\
    \Rightarrow && \frac{1}{4} \l -\ln(1 - v) + \ln(1 + v)\r &= -T
\end{align*}
Since $v = \frac{1}{\sqrt{2}}$

\begin{align*}
    T &= \frac{1}{4} \ln \l \frac{1+ \frac1{\sqrt{2}}}{1 - \frac1{\sqrt{2}}}\r \\
    &= \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r 
\end{align*}

and therefore the total time will be:

\begin{align*}
    & \frac12 \tan^{-1} 1 + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \\
    =& \frac{\pi}{8} +  \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r
\end{align*}