Problem source

In an aerobatics display, Jane  and Karen jump from a great height and go through a period of free fall before opening their parachutes.
While in free fall at speed  $v$, Jane experiences  air resistance $kv$ per unit mass but  Karen, who spread-eagles, experiences air resistance \mbox{$kv + (2k^2/g)v^2$} per unit mass.
Show that Jane's speed can never reach $g/k$. Obtain the corresponding result for Karen.
Jane opens her parachute when  her speed is 
$g/(3{k})$.  Show that she has then been in free fall for time  $k^{-1}\ln (3/2)$. 
 
Karen also opens her parachute  when  her speed is $g/(3{k})$. Find the time she has then been in free fall.

Solution source

Looking at the forces on Jane, $kv < g \Rightarrow v < \frac{g}{k}$. For Karen we have 

\begin{align*}
    kv + (2k^2/g)v^2 &< g\\
    -g^2 + gkv + (2k^2)v^2 &< 0 \\
    (2kv-g)(kv+g) &< 0\\
    \Rightarrow v &< \frac{g}{2k} 
\end{align*}

\begin{align*}
    && \dot{v} &= g - kv \\ 
    \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ 
    \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ 
     && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ 
     && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ 
     && &= \left [-\frac{1}{k} \ln \l g - kv \r  \right ]_0^{g/(3k)} \\ 
     && &= \frac{1}{k} \ln \l g  \r  -  \frac{1}{k} \ln \l \frac{2}{3}g  \r\\ 
     &&&= \frac{1}{k} \ln \l \frac{3}{2} \r
\end{align*}

\begin{align*}
    && \dot{v} &= g - kv - (2k^2/g)v^2 \\ 
    \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ 
    \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ 
     && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ 
     && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ 
     && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ 
     && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ 
     && &= \frac{1}{3k} \ln \l 4 \r
\end{align*}

NB: $\sqrt[3]{4} \approx 1.58 > \frac{3}{2}$ so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.