Answer
a. $x=(140cos22^\circ) t, y=5+(140sin22^\circ) t-16t^2$
b. see graph
c. $48.0\ ft$, $ 1.6\ sec$
d. $3.4\ sec$
e. $437.5\ ft$
Work Step by Step
a. Given $\theta=22^\circ, v_0=140\ ft/s, h=5\ ft$, we can write the position of the projectile as
$x=(140cos22^\circ) t, y=5+(140sin22^\circ) t-16t^2$
where $t$ is the time.
b. We can graph the parametric equations as shown in the figure.
c. We can identify the maximum height as $y_m\approx48.0\ ft$ at time $t\approx1.6\ sec$
d. We can find that the ball will be in the air for $t\approx3.4\ sec$ (at $y=0$, positive solution)
e. We can also identify the range of the ball as $x_m\approx437.5\ ft$ when the ball hits the ground ($y=0$).
