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\begin{document}

\frame{\frametitle{\b 2D Motion - 1 } \bigskip

{\bf Simple projectile motion (1).} \bigskip

Maximum height:
$$ v^0_x = v^0\,\cos(\theta)\,,~~~v^0_y=v^0\,cos(\theta)\,~~~
   v_y=v_y^0-g\,t\,$$
and $v_y=0$ for $y=y_{\max}~\Rightarrow~t_{\max}=v^0_y/g$.
Therefore,
$$ y_{max}=v^0_y\,t_{\max}-g\,t^2_{\max}/2\,.$$
Time at which the maximum height is reached: Already calculated.
\bigskip

{\bf Off the cliff (2):} Insert $t$ from the second equality into
the first:
$$ x=v_0\,t~~{\rm and}~~H=g\,t^2/2\,.        $$
\vskip 20pt }


\frame{\frametitle{\b 2D Motion - 2} \bigskip

\bigskip {\bf Horizontal Distance (3).} \bigskip

First, use 
$$ g\,t_{\max} - v^0=0~~~{\rm and}~~~H=v^0\,t_{max}-\frac{1}{2}\,
   g\,(t_{\max})^2 $$
to express $v_0$ as function of $g$ and $H$. Components:
$$ v^0_x=v^0\,\cos(\phi)\,,~~~v^0_y=v^0\,\sin(\phi)\,.$$
From $g\,t_{\max}^{\phi}-v_y^0=0$ the horizontal distance follows:
$$ d = 2\,t_{\max}^{\phi}\,v_x^0 = \frac{2}{g}\,(v_0)^2\,\cos(\phi)\
   \sin(\phi)\,. $$
Determine $\phi_{\max}$ from the derivative of $d$,
$$ \left. \frac{d\,d}{d\,\phi}\right|_{\phi=\phi_{\max}} = 0\,, $$
and eliminate $v_0$ in terms of $g$ and $H$.
\vskip 20pt }


\frame{\frametitle{\b 2D Motion - 3} \bigskip

\bigskip {\bf Horizontal Range (4).} \bigskip

$$ v_y^0 = \sqrt{2\,h\,g}\,,~~~t=\frac{v_y^0}{g}\,,~~~R=2\,t\,v_x^0\,,
   ~~~v^0=\sqrt{(v_y^0)^2+(v_x^0)^2}\,. $$

\bigskip {\bf Catapult (5).} \bigskip

With $g$, $H$, $D$, and $t$ given:
$$ v_x^0=\frac{D}{t}=\frac{x_f}{t}\,,~~~{\rm and}~~~
   0=H+v^0_y\,t-\frac{1}{2}\,g\,t^2~~\Rightarrow~~v_y^0\,. $$
$\Rightarrow$ angle and speed. Heighest point:
$$ 0 = v_y^0-g\,t_{max}~~~\Rightarrow~~~y_{\max} = H + \cdots\,.$$
\vskip 20pt }


\frame{\frametitle{\b 2D Motion - 5} \bigskip

\bigskip {\bf Circular acceleration (6):} \bigskip

$$T = \frac{2\pi\,r}{v}\,,~~~\omega = \frac{2\pi}{T} 
    = \frac{2\pi\,v}{r} = \frac{v}{r}\,.$$
$$ x=+r\,\cos(\omega\,t)\,,~~~y=+r\,\sin(\omega\,t)\,,$$
$$ v_x=-\omega\,r\,\sin(\omega\,t)\,,~~~
   v_y=+\omega\,r\,\cos(\omega\,t)\,,$$
$$ a_x=-\omega^2\,r\,\cos(\omega\,t)\,,~~~
   a_y=-\omega^2\,r\,\sin(\omega\,t)\,,$$
Therefore,
$$ \vec{r}\,^2=r^2\,,~~\vec{v}\,^2=\omega^2\,r^2\,,~~
   \vec{a}\,^2=\omega^4\,r^2~~\Rightarrow~~
   a=\omega^2\,r=\frac{v^2}{r}\,.$$

\vskip 20pt }



\end{document}
