<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt


{\bf Question}

A particle is projected vertically upward in a constant
gravitational field with an initial speed of $v_0$.  Show that if
there is a retarding force proportional to the square of the
speed, the speed of the particle when it returns to its initial
position is $$\frac{v_0v_T}{\sqrt{v_0^2 + v_T^2}},$$ where $v_T$
is the terminal speed.

\vspace{.25in}

{\bf Answer}

Do this question in 2 parts.

\underline {Upward motion}

\begin{center}
\setlength{\unitlength}{.5in}
\begin{picture}(3,2)
\put(0,0){\vector(0,1){1}}

\put(1,1){\vector(0,-1){0.5}}

\put(1,1){\line(0,-1){1}}

\put(.9,.9){$\bullet$}

\put(0.2,1){\makebox(0,0){$y$}}

\put(1.3,.5){\makebox(0,0){$mg$}}

\put(2,1){\vector(0,-1){0.5}}

\put(2,1){\line(0,-1){.75}}

\put(2.5,0.5){\makebox(0,0){$mk\dot y^2$}}

\end{picture}
\end{center}

Using Newton's 2nd law gives:
\begin{eqnarray*}
 m \ddot y &amp; = &amp; -mk \dot y^2 - mg \\
\frac{dv}{dt} &amp; = &amp; -kv^2 - g {\rm \ \ Put\ }\frac{dv}{dt} = v
\frac{dv}{dy} \\ \int \frac{v \, dv}{kv^2 + Y} &amp; = &amp; -y + C \\
{\rm whence\ }y &amp; =&amp; \frac{1}{2k} \ln \left( \frac{kv_0^2 +
g}{kv^2 +g} \right),
\end{eqnarray*}
where $v_0$ is the initial upwards velocity.


At the highest point $v = 0.$

Height attained $ \ds =  \frac{1}{2k} \ln \left( \frac{k_0^2 +
g}{g} \right)$

\underline  {Downward motion}

\setlength{\unitlength}{.5in}
\begin{picture}(4,2)
\put(1.9,.9){$\bullet$}

\put(2,1){\vector(0,-1){0.5}}

\put(2,1){\vector(0,1){0.5}}

\put(2.3,0.5){\makebox(0,0){$mg$}}

\put(2.2,1.5){\makebox(0,0)[l]{$mk \dot y^2$}}
\end{picture}


Using Newton's 2nd law gives:

\begin{eqnarray*} \ddot y &amp; = &amp; k \dot y^2 - g
\\ y &amp; = &amp; \frac{1}{2k} \ln \left( \frac{g - kv^2}{g} \right) +
\frac{1}{2k} \ln \left( \frac{kv_0^2 + g}{g} \right),
\hspace{.1in} (*)\end{eqnarray*}

where $v = 0$ at the highest point.

Solving for $v$ gives $v  =  \sqrt{\frac{\frac{g}{k}v_0^2}{v_0^2 +
\frac{g}{k}}}$


Terminal velocity occurs when there is no net force on the mass
while the particle is falling i.e. $ \ds mkv_t^2 = mg \Rightarrow
v_t = \sqrt{\frac{g}{k}}$.   $$ \textrm{Therefore \ \ } v =
\frac{v_0v_t}{\sqrt{v_0^2 + v_t^2}}$$


\end{document}
</pre></body></html>