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\begin{document}

\newpage
% 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.
\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\HOW DEEP IS THE WELL?}






\markboth{How Deep is the Well?}{How Deep is the Well?}






\author{Brian Winkel, Director
SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
}

\makeStitlePDFLaTex



\begin{abstract}
We drop a pebble in a deep well.  Given the time elapsed from release of the pebble until we hear the splash determine the depth of the well.
\end{abstract}


\section*{SCENARIO DESCRIPTION}



\begin{enumerate}
\item You drop a pebble into a  well.  Ten seconds later, you hear the  sound of the pebble hitting the water.  How deep is the well?  For what part of the 10 seconds was the pebble falling?  How long did it take the sound to travel back up the well?  What assumptions did you make in solving the problem?
\item Suppose another well is just that depth at which the time it takes your pebble to hit the water is the same as the time it takes the sound of it hitting the water to return to the top.  How deep is this well?
\item Suppose another well is just that depth at which the time it takes your pebble to hit bottom is 15  times greater than the time it takes the sound of it hitting to return to the top.  How deep is this well?
\end{enumerate}



\teacher{
\section*{NOTES FOR TEACHERS}

The pebble in free fall downward can be modeled  using  Newton's Second Law of Motion which says that the force of mass times the acceleration of a moving body is the sum of all the external forces acting on the body.  Since the only force acting on the falling pebble is presumed to be gravity (we shall ignore resistance here) then we have the following differential equation (\ref{eq:original}) describing the fall:

\vspace*{-.4 true cm}
\begin{equation}\label{eq:original}
m\cdot y''(t) = - m \cdot g \, ,
\end{equation}
where $m$ is the mass of the pebble, $g$ is the acceleration due to 
gravity (here $g = 9.81{\rm m/sec}^2$), and $y(t)$ is the displacement in meters at time $t$ in seconds from release. $y(0) = 0$ is the initial displacement and distance of the fall is measured negatively downward. 

The solution to (\ref{eq:original}) is


\vspace*{-.4 true cm}
\begin{equation}\label{eq:sol}
y(t) = -\frac{g}{2}t^2 + v_0 t + y_0\, .
\end{equation}


Since the pebble is dropped, not thrown, the initial velocity, $y'(0) = v_0$, is  zero, so the linear term, $v_0 t$, is zero.  Further, assume that the pebble is dropped from ground level.  This makes the constant term, $y_0$, also zero.  Thus,  the position of the falling pebble as a function of time is given by 

\vspace*{-.5 true cm}
\begin{equation}
\label{eq:s}
y(t) = -\frac{g}{2}t^2 = -\frac{9.81}{2}t^2\, .
\end{equation}
Note that this position function is negative.  It should be, since the pebble is going down a well.  



 The distance of the sound travelling back up the well is a linear function 
\begin{equation}
\label{eq:d}d(t) = 340t\, ,
\end{equation}
 where $d(t)$ is the distance in meters, $t$ is in seconds, and the 340 represents the approximate speed of sound measured in meters per second.

 


However, this problem is about time, not distance, so solve both (\ref{eq:s}) and (\ref{eq:d})   for $t$. Then, add these two times and set that sum equal to 10 seconds. When solving for time for each process (1) falling and (2) sound travel let $s$ be the depth of the well in meters. 

\begin{equation}\label{eq:tot}
{\rm Total}(s) = {\rm Time\,\, falling\,\, down} + {\rm Time\,\, sound\,\, up} = \sqrt{\frac{s}{4.905}} + s/340 = 10\, .
\end{equation}

We can solve (\ref{eq:tot}) to obtain $s = 385.562$ m. Upon substitution we then find the time for the pebble to fall down the well to be  8.86599 s while the time for the sound to travel up is 1.13401 s and these add up to 10.0 s!

Now we ask how deep must the well be for the time of the fall down to be the same as the time of the sound to return after the splash?


We get two solutions.  The first is trivial.  If the well has no depth, the pebble takes no time to fall, and the sound takes no time to return.  The second is more interesting.  A depth of 23,567.8 meters is the other solution.  This is a very deep well indeed, almost 15 miles!!! Obviously, this is a head game question only and not really possible.

So we change our question slightly to finding the well depth for which the time it takes the pebble to hit bottom to be 15 times greater than the time it takes the sound of it hitting to return to the top. Here too we get the trivial solution of depth zero, but also a depth of 104.746 m which, while big, is not outrageous.

\subsection*{ISSUES RELATED TO THE PROBLEM}



This problem combines two types of motion, each describe a different family of functions.  The pebble in free fall conforms to a differential equation, while the sound moving back up the well conforms to a linear equation, indeed, used $D = R*T$ where $D$ is distance sound travels; $R$ is the speed of sound; and $T$ is the time the sound travels. 

 
\subsubsection*{Prerequisites}

The students need to be able to build a differential equation model for falling body (without resistance) and the distance = rate * time formula.  They will also have to know how recognize that the total time elapsed consists of the time it takes for the rock to fall added to the time it takes the sound to reach the well top.

\subsubsection*{Time allotment - time management}

This problem can be done in one class period, even half a period or 20 minutes if you let students work in groups and help each other realize the importance of adding up the times.  
\subsubsection*{Expectations} 

Students will sum the times and arrive at a complex function which when set to the observed time will need to be solved, perhaps with the help of technology, say a root finder.

\subsubsection*{Extensions}
 


The problem suggests several descriptions of wells which should give the teacher a number of further extensions.  Indeed, collecting data on a given well might help.  Of course, one could get into the differential equations describing the falling pebble as it meets resistance in its fall in the well as well!



\section*{ACKNOWLEDGEMENT}


The production of this material was supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849 from 1993-1996 entitled Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. This material was produced by Dave Horn a high school mathematics teacher from Michigan City IN as part of a faculty developer workshop with  the original full title, {\it How Deep is the Well\/}.  
}


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