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\begin{document}

\noindent
{\bf Relevant Formulas and Data}

\headline
$\ds A=\frac{1}{2} \theta r^2 \hskip.25in$ Area of parabolic section $ \ds = \frac{4}{3} \times$ Area of
inscribed triangle.

\bigno
$\ds \frac{a_M^3}{T_M^2} = \frac{a_J^3}{T_J^2}$

\bigno
$\ds y = Ax^2 + Bx +C \hskip.25in \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \hskip.25in a = \frac{k}{2}$
\, and \, $\ds b=\sqrt{a^2 - e^2}.$

\bigno
$r=a (1-\varepsilon \, \cos \, \beta) \hskip.25in \tan \ds \frac{\alpha}{2} = \sqrt{\frac{1+ \varepsilon}{1-
\varepsilon}} \ \tan \frac{\beta}{2} \hskip.25in \beta - \varepsilon \sin \beta = \frac{2 \pi t}{T}$

\bigno
$\ds \frac{2 \pi t}{T} + \varepsilon \ ( \ \ )$

\vskip.5in 

%\begin{center}
\begin{tabular}{|c|c|c|c|c|c|} \hline
\multicolumn{6}{|c|}{\bfseries Orbital Data of Planets}\\ \hline 
Planet & Semi-major & Period of the & Astronomical  & Angle of orbital & Average speed \\
& axis in AUs & orbit in years & Eccentricity & plane with that & in miles/sec \\
& & & & of the Earth & \\
\hline
Mercury & 0.3871 & 0.2408 & 0.2056 & $7.00^{\circ}$ & 29.6 \\
\hline
Venus & 0.7233 & 0.6152 & 0.0068 & $3.39^{\circ}$ & 21.7 \\
\hline
Earth & 1.0000 & 1.0000 & 0.0167 & $0.00^{\circ}$ & 18.5 \\
\hline
Mars & 1.5237 & 1.8809 & 0.0934 & $1.85^{\circ}$ & 15.0 \\
\hline
Jupiter & 5.2028 & 11.8622 & 0.0484 & $1.31^{\circ}$ & 8.1 \\
\hline
Saturn & 9.5388 & 29.4577 & 0.0557 & $2.49^{\circ}$ & 6.0\\
\hline
\end{tabular}
%\end{center}

\vskip.5in

\noindent
$\ds \int_a^b  f(x) \ dx = F (b) - F(a) \hskip.25in A=\int_a^b \pi f \, (x)^2 \, dx \hskip.25in L =
\int_a^b \sqrt{1+f' (x)^2} \ dx$

\bigno
$\ds \int_a^b cx^r \ dx = \ds \frac{c}{r+1} x^{r+1} |_a^b = \frac{c}{r+1} b^{r+1} - \frac{c}{r+1}
a^{r+1}$

\bigno
$x' (t) = v_0 \cos \varphi_0 \ $ \, and \, $x(t) = \left(v_0 \cos \varphi_0 \right)t,$

\bigno
$y' (t) = -gt + v_0 \sin \varphi_0$ \, and \, $\ds y(t) = -\frac{g}{2} t^2 + \left(v_0 \sin \varphi_0 \right)
t+y_0$

\bigno
$\ds y= \left(\frac{-g}{2v_0^2 \cos^2 \varphi_0}\right) x^2 + \left(\tan \ \varphi_0 \right) x+y_0$

\bigno
$\ds \frac{1}{2g} v_0^2 \sin^2 \varphi_0 + y_0$ (maximal height attained by the projectile)

\bigno
$\ds t_{\mbox{imp}} = \frac{v_0 \sin \varphi_0 + \sqrt{v_0^2 \sin^2 \varphi_0 + 2gy_0}}{g}$ (time of
impact)

\bigno
$\ds R= \frac{v_0^2}{2g} \sin (2 \varphi_0) + \frac{v_0}{g} \sqrt{\frac{v_0^2}{4} \sin^2 (2 \varphi_0)
+ 2gy_0 \cos^2 \varphi_0}$ \, (range)

\bigno
$R_{\mbox{max}} = \ds \frac{v_0^2}{2g} + \frac{v_0}{g} \sqrt{\frac{v_0^2}{4} + gy_0}$ (maximal
range)

\bigno
$\ds v(t) = \sqrt{v_0^2 + g^2 t^2 -2 g \left(v_0 \sin \varphi_0 \right) t}$





\end{document}

