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\documentclass{simiode}
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\newcommand{\teacher}[1]{\ifnum\TeacherEdition=1 #1 \fi}
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% the document. See this file for examples. This has been done for you in the document below.
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\begin{document}

\newpage
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% You only need to change the NAME OF SCENARIO in the next line and delete the footnote.
% See above for instructions on toggling from Teacher Version to Student Version.
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\title{\ifnum\TeacherEdition=1 TEACHER VERSION
%\footnote{DELETE THIS FOOTNOTE BEFORE SUBMISSION.  In the \TeX \, source code there is a toggle/option to produce a Teacher Version or a Student Version by ``carving'' out with  $\backslash$teacher\{\} that material which should only appear in the Teacher version. See the source code.}
\else STUDENT VERSION \fi\\A Shot in the Water}
%\footnote{Inspired by the article ``Slings, Bullets, Blow-up, and Linearity'' in SIAM News, vol. 51, Issue 1, January/February 2018.}}



\teacher{
    \keywords{projectile motion, friction, air resistance}

%\tagwords{Here you can enter a wider set of words which serve as tags (as do keywords) in the SIMIODE web site. Keep this list to ten words. The Editor may suggest/change/add/delete some for more refinement as necessary.}
\tagwords{terminal velocity, water resistance, reduction of order, phase line portrait,  second order, first order, phase portrait, differential equation}
}

\markboth{A Shot in the Water}{A Shot in the Water}


\author{Kurt Bryan\\
Rose-Hulman Institute of Technology\\
Terre Haute IN 47840 USA\\
kurt.bryan@rose-hulman.edu
}

\makeStitlePDFLaTex


\begin{abstract}
This project uses Newton's Second Law of Motion in conjunction with a quadratic model for the resistance experienced by a bullet moving through water
to analyze a classic action movie scene: Do bullets moving through water slow as dramatically as depicted in the movies, so that someone a few feet underwater is safe?
The mathematics involved reduces to a standard first order ``air resistance'' ordinary differential equation obtained in most introductory classes on
differential equations.
A computer algebra system can be helpful for performing computations, but is not essential. This project can be done in class, or as a homework assignment.
\end{abstract}


\section*{SCENARIO DESCRIPTION}

It's a classic scene from an action movie: our hero is fleeing the bad guys and dives over the side of a boat to escape, then begins to swim away underwater. The villains
draw guns and proceed to fire at the hero, but the view beneath the surface shows the bullets slowing dramatically as they enter, quickly rendering them harmless. But is this realistic?

Let's model the situation. Let $y$ denote the vertical coordinate, with $y>0$ corresponding to increasing depth and $y=0$ the water surface.
We'll use $y(t)$ to denote the position of the bullet at time $t$, and we'll assume the bullet's motion is purely vertical. From Newton's Second Law which says that  ``the mass times the acceleration applied to that mass is equal to the sum of all the external forces applied to that mass'' we have
\begin{equation}
\label{newton2}
my''(t) = F
\end{equation}
where $m$ is the mass of the bullet, $y''$ is the acceleration of the bullet, and $F$ denotes the net force acting on the bullet, that is, $F$ is the sum of all the forces acting on the bullet. One component of the force is $F_g = mg$, where $g>0$ denotes gravitational
acceleration. The other dominant force is the resistance presented by the bullet's motion through the water.  We'll take a classic quadratic model for this resistance, in the form
$F_w = k(y'(t))^2$
for some constant $k$. We're interested in the case in which the bullet is moving downward, so $F_w$ should point upward (the negative direction), and hence $k$ would be negative.
Better yet, let us take
\begin{equation}
\label{resis}
F_w = -k (y'(t))^2
\end{equation}
with $k>0$. All in all, our model for the motion of the bullet is then obtained from $F = F_g + F_w$, which in (\ref{newton2}) yields
\begin{equation}
\label{ode}
my''(t) = mg - k (y'(t))^2.
\end{equation}
Let's assume that the bullet enters the water at time $t=0$, so $y(0)=0$, and that $y'(t) = v_0>0$ for some (downward) initial velocity $v_0$.\\

%\section*{Submission Version}
\noindent {\bf Analysis}\\

First, although the differential equation in (\ref{ode}) is technically second order, we can reduce it to first order by letting $v(t) = y'(t)$. In this case the equation becomes
\begin{equation}
\label{odev}
m v'(t) = mg - k (v(t))^2
\end{equation}
a first order equation with initial condition $v(0)=v_0$. This is the equation we will work with for now.

Let's assume that bullet has a mass of $55$ ``grains'' (a typical unit and mass for a rifle bullet), which corresponds to $m=3.563\times 10^{-3}$ kg. We'll take $g=9.81$ meters per second squared. A typical modern high velocity rifle has a muzzle velocity in excess of $1000$ meters per second, so let's go with $v_0=1000$.

The only unknown parameter at the moment is $k$. One way we can get an estimate of $k$ is as follows:\\\ \\
{\bf Exercise 1:} Sketch a phase portrait for (\ref{odev}); you can restrict your attention to the region $v>0$ (bullet moving downward). What is the fixed point for this ODE in terms of $m,g,$ and $k$, and what is the fixed point's physical meaning?\\\ \\
{\bf Exercise 2:} Suppose, for argument's sake, that the bullet would fall at a terminal velocity of $1$ meter per second if dropped in the water. Use this in conjunction with the answer to Exercise 2 to estimate $k$. (It would be better to do an experiment to estimate $k$; go for it, if you have the means, and let me know the answer).\\

Now that we have all the vital parameters, we can solve (\ref{odev}). The solution can be a bit messy and hard to interpret, even with a computer algebra system. To simplify matters and more clearly see the structure of the problem, let's first define a new function $w(t)$ as
\begin{equation}
\label{wdef}
w(t) = v(t)/g.
\end{equation}
{\bf Exercise 3:} Show that $w$ satisfies the differential equation
\begin{equation}
\label{wde}
w'(t) = 1 - B^2 (w(t))^2
\end{equation}
where $B = \sqrt{kg/m}$ (note $kg/m>0$, so $B$ is real). Also, $w(0)=v_0/g>0$. It's probably easier to leave $m,g,k,$ and $v_0$ as unspecified parameters for now, rather than filling in numerical values.\\\ \\
{\bf Exercise 4:} Solve (\ref{wde}) using separation of variables to show that
\begin{equation}
\label{wsol}
w(t) = {\frac {1}{B}\tanh \left( B t+{\rm arctanh} \left({
\frac{Bv_0}{g}}\right) \right) }
\end{equation}
You may find it helpful to note that
$$
\int \frac{du}{1-B^2 u^2} = B\;{\rm arctanh}(B u) + C.
$$

After solving for $w(t)$ as in (\ref{wsol}), show that
\begin{equation}
\label{vsol}
v(t) = {\frac {g}{B}\tanh \left( Bt+{\rm arctanh} \left({
\frac{Bv_0}{g}}\right) \right) }.
\end{equation}
\ \\
{\bf Exercise 5:} Let's suppose the bullet is harmless once $v(t) \leq 10$ meters per second. Use (\ref{vsol}) with the given $m,g,k,$ and $v_0$ to determine that time $t^*$ when $v(t^*)=10$
(you may have to solve numerically; it might be helpful to plot $v(t)$).  Then compute the distance traveled by the bullet from $t=0$ (when the bullet enters
the water at $y=0$) until $t=t^*$. How deep must our hero be to escape serious injury?\\\ \\
{\bf Exercise 6:} Suppose our estimate of the terminal velocity of the bullet is wrong---what if it descends at a terminal velocity of $2$ meters per second---how much difference does that make?
%What if the resistance experienced by the bullet is linear with respect to $v(t)$---how much difference does that make?
%{\bf Exercise 7:} Ignoring relativistic effects, suppose $v_0=-10^8$ (about $1/3$ the speed of light!) How deep must our hero be now?
\teacher{
\section*{NOTES FOR TEACHERS}

With an assumed terminal velocity of $1$ meter per second we find in Exercise 2 that $k\approx 0.03496$ (units: kg per meter after simplifying). This follows from
the fact (established in Exercise 1) that the terminal velocity is $\sqrt{mg/k}$. If we use a terminal velocity of $2$ meters per second as in Exercise 6 we find that
$k\approx 0.00874$. In Exercise 5 we find with $k=0.03496$ that the bullet has slowed to a speed of $10$ meters per second after only $0.0101$ seconds, and travels a distance of only $0.47$ meters in this time span, less than $2$ feet! But with $k=0.00874$ the bullet takes about $0.0409$ seconds to slow to a harmless speed, and has traveled about $1.89$ meters.
So it makes a big difference---we'd want a good estimate of the terminal velocity/resistance experienced by the bullet before assuming our hero is safe.

This project can be done without technology, but my students have Maple and found it beneficial to plot things as they went along. Also, the integration of $v(t)$ in Exercise
5 is a bit cumbersome.
I usually assign this project as homework after discussing basic motion models in class, and the notion of phase line portraits for autonomous systems. The students have never consistently gotten stuck on any particular part of the project, although they frequently make the mistake of taking various velocities (e.g., the terminal velocity, or the entry velocity of the bullet) to be negative, when in fact our coordinate system has these as positive.}

\section*{ACKNOWLEDGEMENTS}

This project was inspired by the article \cite{siam}. See also the video \cite{youtube1} for a rather dramatic demonstration of the conclusions of this analysis!

\begin{thebibliography}{99}
\bibitem{siam} Levi, M. 2018.  Slings, Bullets, Blow-up, and Linearity.  {\it SIAM News\/}. 51(1):  January/February.
\bibitem{youtube1} NRK Viten. 2016. Fires weapon under water - with his own life on the line. \url{https://www.youtube.com/watch?v=tzm_yyl13yo}. Accessed April 5, 2018.
\end{thebibliography}




\end{document}
