A projectile is launched from ground level at an angle of $\theta$ with the horizontal. The initial velocity is $v _ { 0 }$ feet per second and the path of the projectile is modeled by the parametric equations
$x = \left( v _ { 0 } \cos \theta \right) t$ and $y = \left( v _ { 0 } \sin \theta \right) t - 16 t ^ { 2 }$.
Use a graphing utility to graph the paths of a projectile launched from ground level with the values given for $\theta$ and $v _ { 0 }$. Use the graph to approximate the maximum height and range of the projectile to the nearest foot.
$\theta = 50 ^ { \circ } , \quad v _ { 0 } = 104$ feet per second
A) maximum height $\approx 12$ feet range $\approx 171$ feet
B) maximum height $\approx 171$ feet range $\approx 12$ feet
C) maximum height $\approx 99$ feet range $\approx 333$ feet
D) maximum height $\approx 333$ feet range $\approx 99$ feet
E) maximum height $\approx 38$ feet range $\approx 8$ feet

Question

Quizplus · Accepted Answer

Using a graphing utility, we can plot the parametric equations $x = \left( v _ { 0 } \cos \theta \right) t$ and $y = \left( v _ { 0 } \sin \theta \right) t - 16 t ^ { 2 }$ with $\theta = 50 ^ { \circ }$ and $v_0 = 104$ to obtain the following graph:

\begin{center}
\begin{tikzpicture}[scale=0.8]
\begin{axis}[
    xmin=0,xmax=350,
    ymin=0,ymax=140,
    axis lines=middle,
    xlabel={$x$},
    ylabel={$y$},
    x label style={at={(current axis.right of origin)},anchor=north},
    y label style={at={(current axis.above origin)},anchor=east},
    grid=major,
    grid style={dashed, gray!30},
]
\addplot[domain=0:20.3,black,-] ({104*cos(50)*x},{104*sin(50)*x-16*x^2}) node[midway, sloped, above]{$y = -16x^2+1067x$};
\end{axis}
\end{tikzpicture}
\end{center}

We can see from the graph that the projectile reaches its maximum height when the slope of the tangent line is equal to zero, or when the $y$-coordinate is maximized. This occurs at approximately $x=33$ feet, which corresponds to a $y$-coordinate of about $99$ feet. Therefore, the maximum height is approximately $99$ feet.

To find the range, we need to find the $x$-coordinate of the point where the projectile hits the ground. This occurs when $y=0$, so we can solve $0 = \left( v _ { 0 } \sin \theta \right) t - 16 t ^ { 2 }$ for $t$ to find
$$t = \frac{v_0\sin\theta}{16} = \frac{104\sin 50^{\circ}}{16} \approx 4.89.$$
Then, we can find the $x$-coordinate by plugging in $t=4.89$ to the equation $x = \left( v _ { 0 } \cos \theta \right) t$, giving us
$$x = 104\cos 50^{\circ} \cdot 4.89 \approx 333.$$
Therefore, the range is approximately $333$ feet.

Thus, the best choice is C, maximum height $\approx 99$ feet and range $\approx 333$ feet.

A Projectile Is Launched from Ground Level at an Angle θ\thetaθ

A Projectile Is Launched from Ground Level at an Angle $\theta$