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% Semester
\newcommand{\semester}{Fall 2003}
% Name of the class.
\newcommand{\class}{Math 235--Calculus Laboratory}
% Title.
\newcommand{\labtitle}{
 Which is Faster, Going Up or Coming Down?
  }

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\noindent \semester  \hfill \class
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{\bf \large \em \labtitle}\footnote{taken from Calculus, 4th ed. Stewart}
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Suppose you throw a ball into the air.  Do you think it takes longer to
reach its maximum height or to fall back to Earth from its maximum height?
We will solve the problem in this project but, before getting started,
think about that situation and make a guess based on your physical
intuition.

\begin{enumerate}

 \item A ball with mass $m$ is projected vertiaclly upward from Earth's
 surface with a positive initial velocity $v_0$.  We assume the forces
 acting on the ball are the force of gravity and a retarding force of air
 resistance with direction opposite to the direction of motion and with
 magnitude $p|v(t)|$, where $p$ is a positive constant and $v(t)$ is
 the velocity of the ball at time $t$.  In both the ascent and descent, the
 total force acting on the ball is $-pv-mg$.  (During the ascent, $v(t)$ is
 positive and the resistence acts downward; during descent, $v(t)$ is
 negative and the resistance acts upward.)  So, by Newton's Second Law, the
 equation of motion is $$mv'=-pv-mg.$$  Solve this differential equation. 
 
 \item Determine $y(t)$, the height of the ball at time $t$ (until
 it hits the ground).

 \item \label{maxht} Let $t_1$ be the time that the ball takes to reach its
 maximum height.  Show that $$t_1 = \frac{m}{p} \ln\left(
 \frac{mg+pv_0}{mg} \right).$$  Find this time for a ball with mass 1 kg
 and initial velocity 20 m/s.  Assume the air resistance is 1/10 of the
 speed.

 \item Let $t_2$ be the time at which the ball falls back to Earth.  For
 the particular ball in Problem \ref{maxht}, estimate $t_2$ by using a
 graph of the height function $y(t)$.  Which is faster, going up or coming
 down?

 \item In general, it's not easy to find $t_2$ because it's impossible to
 solve the equation $y(t)=0$ explicitly.  We can, however, use an indirect
 method to determine whether ascent or descent is faster; we determine
 whether $y(2t_1)$ is positive or negative.

 Show that $$y(2t_1) = \frac{m^2g}{p^2}\left( x - \frac{1}{x} - 2 \ln x
 \right) $$ where $x=\exp(pt_1/m)$.  The show that $x>1$ and the function
 $$f(x)=x-\frac{1}{x} - 2\ln x$$ is increasing for $x>1$.  Use this result
 to decide whether $y(2t_1)$ is positive or negative.  What can you
 conclude?  Is the ascent or descent faster?

\end{enumerate}


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