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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Applications of the derivative.}

\begin{example}
A ball is thrown vertically upward with a velocity of \( 80\, ft/sec \), and
its position \( t \) seconds later is \( s(t)\, :=80t-16t^{2} \) feet up.
How high does it get, and how fast is it moving when it hits the ground again?
\end{example}

\subsubsection*{Solution:}

The key thing for the first question is to note that the ball reaches velocity
\( 0 \) when it is at its highest point. Then, solve \begin{eqnarray*}
0 & = & v(t)\\
 & = & 80-32t,
\end{eqnarray*}
which implies that \( t=80/32=5/2 \). The maximum height is then \begin{eqnarray*}
s(5/2) & = & 80(5/2)-16(5/2)^{2}\\
 & = & 200-100\\
 & = & 100.
\end{eqnarray*}
For the second question, solve for \( s=0 \): \begin{eqnarray*}
0 & = & s(t)\\
 & = & 80t-16t^{2}\\
 & = & 16t(5-t),
\end{eqnarray*}
so \( t=0 \) or \( t=5 \). Since the motion started when \( t=0 \), the ball
had to hit the ground when \( t=5 \). To then answer the question, the velocity
is \( v(5)=80-32\cdot 5=-80. \) This makes sense, since it really means \( 80\, ft/sec \)
\textit{downward}.
\end{document}
