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%TCIDATA{Created=Tue Aug 20 16:37:22 1996}
%TCIDATA{LastRevised=Tue Jan 21 16:32:05 1997}
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\input{tcilatex}
\begin{document}


\begin{abstract}
A Laboratory report created with Scientific Notebook
\end{abstract}

\author{A. U. Thor}
\title{Lab Report}
\date{The Date }
\maketitle

\section{Projectile Motion}

\begin{center}
\textbf{Brian Palmer and Steve Nava}

Wednesday, January 1, 1997
\end{center}


Working together, we completed Lab \#5 as given to us in class. Summarizing,
Steve held the ball at what was determined by measurement to be about $1.895$
meters from the ground. Then he threw it, releasing the ball from his hand
at that spot. It struck the ground $2.56$ seconds later, at a point $43.92$
meters from the initial throwing position.

The data was then available for analysis. Although it was performed in three
dimensions, there are only two colinear points which were recorded; thus the
third dimension (along the $z$-axis) can be ignored (this lab assumes that
the force along the other axes applied by the wind blowing at the time was
negligible). Then each of the two remaining dimensions, along the $x$ and $y$
axes, to use traditional terminology, may be analyzed as if the other were
not present. For the initial analysis, the kinematics equation, where $%
\Delta x$ is the position displacement, $v_{0}$ is the initial velocity, $a$
is the acceleration undergone by the object, and $t$ is the time through
which the analysis is concerned, 
\begin{equation}
\Delta x=v_{0}t+\frac{1}{2}at^{2}  \tag{1}  \label{kinematics}
\end{equation}
shall be used extensively.

When dealing with the $x$-component of the system, the relationship given in
equation 1 can be explicated as follows 
\[
\Delta x_{x}=v_{0x}t+\frac{1}{2}a_{x}t^{2}
\]
Since after the initial force applied by the throwing action, there is no
force acting on the tennis ball, the relationship $a=\frac{F}{m}$ for a
force $F$ acting upon a mass $m$ shows that $a=\frac{0}{m}=0$ $\limfunc{m}/%
\limfunc{s}^{2}$. Simplifying the relationship further, 
\[
\Delta x_{x}=v_{0x}t
\]
The displacement and time are known, allowing the initial velocity to be
discovered 
\begin{equation}
v_{0x}=\frac{\Delta x_{x}}{t}=\frac{43.92\limfunc{m}}{2.56\limfunc{s}}=17.156%
\limfunc{m}/\limfunc{s}\approx 17.16\limfunc{m}/\limfunc{s}
\end{equation}

The unknown value of the initial velocity's $y$ component can be discovered
similarly. However, for this, a force is known to be acting upon the tennis
ball: gravity. Gravity can, for any mass $m$ such that the mass of the Earth 
$M_E\gg m$, the acceleration due to gravity on the object may be
approximated as the constant $g$, about $9.8006\limfunc{m}/\limfunc{s}^2$.
Thus, 
\begin{eqnarray}
\Delta x_y &=&v_{0y}t+\frac 12at^2  \label{yexpress} \\
v_{oy}t &=&\Delta x_y-\frac 12at^2 \\
v_{oy} &=&\frac{\Delta x_y}t-\frac 12at  \nonumber \\
v_{oy} &\approx &\frac{\Delta x_y}t-\frac 12gt
\end{eqnarray}
The time $t$ is the same as that for the $x$-component; that is, it is the
time until the ball hit the ground. Substituting in the values, and using
the definition that $\Delta h=h_f-h_i$, 
\[
v_{0y}=\frac{x_{yf}-x_{yi}}t-\frac 12gt=\frac{0-1.895\limfunc{m}}{2.56%
\limfunc{s}}-\frac 12\left( 9.8\limfunc{m}/\limfunc{s^2}\right) \left( 2.56%
\limfunc{s}\right) \approx -0.74\limfunc{m}/\limfunc{s}+12.54\limfunc{m}/%
\limfunc{s}=13.284\limfunc{m}/\limfunc{s} 
\]

At last, the $x$ and $y$ components of the initial velocity have been
calculated. Using some of the definitions and theorems associated with
vectors, the characteristics of the vector as a whole can be calculated.
First, the magnitude of the initial velocity vector, $\left| \vec{v}\right| $%
, can be calculated as 
\[
\left| \vec{v}_{0}\right| =\sqrt{v_{0x}^{2}+v_{0y}^{2}}\approx \sqrt{\left(
17.16\limfunc{m}/\limfunc{s}\right) ^{2}+\left( 13.284\limfunc{m}/\limfunc{s}%
\right) ^{2}}\approx 21.7\limfunc{m}/\limfunc{s}
\]
The direction $\theta $ can also be determined, using the definition 
\[
\theta =\arctan \left( \frac{v_{0y}}{v_{0x}}\right) \approx 37.74\unit{%
\UNICODE{0xb0}}
\]

In general form, the calculations above for the vertical velocity can be
used to derive the equation of vertical velocity expressed as a function of
time, as well as the fact that the velocity $v_{y}=0$ $\limfunc{m}/\limfunc{s%
}$ at the vertex of the flight. Using another fundamental equation of
kinematics, 
\begin{eqnarray*}
v_{y} &=&v_{0y}^{2}+2\vec{a}\Delta x_{y} \\
v_{y} &=&\sqrt{v_{0y}^{2}+2\vec{a}\Delta x_{y}}
\end{eqnarray*}
can be figured as describing the vertical velocity $v_{y}$. The highest
point (or rather, the maximum $\Delta x_{y}$) that shall be encountered in
the trajectory can be discovered using the principle derived from min-max
functions of calculus that at the vertex of the parabola, the velocity $v_{y}
$ $=0$ $\limfunc{m}/\limfunc{s}$. Rearranging the equation, 
\[
\Delta x_{y}=\frac{v_{y}^{2}-v_{0y}^{2}}{2\vec{a}}
\]
Substiting in known values, 
\begin{eqnarray*}
\Delta x_{y} &=&\frac{\left( 0\limfunc{m}/\limfunc{s}\right) ^{2}-\left(
13.28\limfunc{m}/\limfunc{s}\right) ^{2}}{2\left( -9.8\limfunc{m}/\limfunc{s}%
\right) } \\
\Delta x_{y} &=&\frac{\left( 13.28\limfunc{m}/\limfunc{s}\right) ^{2}}{19.6%
\limfunc{m}/\limfunc{s}}=\frac{176.36\limfunc{m}/\limfunc{s}}{19.6\limfunc{m}%
/\limfunc{s}} \\
\Delta x_{y} &=&8.998\limfunc{m}
\end{eqnarray*}
Since this is the difference in the $x_{y}$ values, the maximum height of
the tennis ball is this value plus its initial value, or $8.998\limfunc{m}%
+1.895\limfunc{m}=10.893\limfunc{m}$.

\end{document}
