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\large{\textbf{Math 2250 Homework \#3}}
\end{center}
\bigskip
This homework assignment covers the robot range lab. The setup for the lab is that we have built a robot to throw a ball by attaching the ball to a rotating arm using an electromagnet. The arm has length 1 m (for now) and rotates around a center at $(0,0)$ at an (angular) speed of 1 radian per second (for now). The position of the arm is then given by
\begin{equation*}
x_{\text{arm}}(\theta) = \cos \theta. \qquad y_{\text{arm}}(\theta) = \sin \theta.
\end{equation*}
At some angle $\theta_0$, we release the ball. We assume that this is time $t=0$ and that the ball then follows a parabolic path given by some functions
\begin{equation*}
x_{\text{ball}}(t) = a t^2 + b t + c. \qquad y_{\text{ball}}(t) = p t^2 + q t + r.
\end{equation*}
This is shown in the pictures
\begin{center}
\begin{tabular}{cccc}
\includegraphics[width=2in,trim = 0.75in 1in 0.75in 1in]{robot_throw_1} &
\includegraphics[width=2in,trim = 0.75in 1in 0.75in 1in]{robot_throw_4} &
\includegraphics[width=2in,trim = 0.75in 1in 0.75in 1in]{robot_throw_3} \\
ball starts at $\theta=0$ \dots &
\dots released at $\theta=\theta_0$ \dots &
\dots and flies along parabola. \\
\end{tabular}
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\section{Problems}
\begin{problems}
\item Find $x_{\text{ball}}(t)$ and $y_{\text{ball}}(t)$ for a given $\theta_0$. Your answer should involve various trig functions of $\theta_0$. Keep in mind that the position and velocity of the \textbf{arm} at $\theta = \theta_0$ are the same as the position and velocity of the \textbf{ball} at time $t = 0$. Use units of meters and seconds so that the acceleration of gravity is $-9.8 m/s^2$.
\bigskip
\noindent\textbf{Note:} The answer we arrived at in class was
\begin{equation*}
x_{\text{ball}}(t) = -\sin \theta_0 t + \cos \theta_0. \qquad
y_{\text{ball}}(t) = -4.9 t^2 + \cos \theta_0 t + \sin \theta_0.
\end{equation*}
\item For a fixed $\theta_0$, find the maximum value of $y_{\text{ball}}(t)$. The answer is a new function $H(\theta_0)$ which tells us how \textbf{high} the robot will throw the ball if we release at angle $\theta_0$. This is a max/min problem where we take the derivative with respect to $t$.
\item Find the maximum value of $H(\theta_0)$ as $\theta_0$ varies from $0$ to $2\pi$. This tells us the release angle which allows the robot to throw the ball \emph{highest}.
\item Suppose the ground is located $y=-1.25$ (that is, we locate the motor on a stand so that it is 1.25 meters above the ground). For a fixed $\theta_0$ use your $y_{\text{ball}}(t)$ function to figure out \emph{when} the ball hits the ground. Use that time and your $x_{\text{ball}}(t)$ function to figure out \emph{where} the ball hits the ground. The answer is a new function $R(\theta_0)$ which tells us how \textbf{far} the robot will throw the ball if we release at angle $\theta_0$.
\item Find the maximum value of $R(\theta_0)$ as $\theta_0$ varies from $0$ to $2\pi$. This tells us the release angle which allows the robot to throw the ball \emph{farthest}.
\item Suppose we want to hit a target at position $(x_{\text{target}},y_{\text{target}})$. Use your $x_{\text{ball}}(t)$ and $y_{\text{ball}}(t)$ functions to solve for the release angle (or angles!) which will cause the ball to pass through $(x_{\text{target}},y_{\text{target}})$. You won't be able to solve the equations that you set up for any $(x_{\text{target}},y_{\text{target}})$: what does this mean?
\end{problems}
\end{document}