Answer
a. $\begin{cases} x=(150\ cos35^\circ) t\\ y=3+(150\ sin35^\circ) t -16t^2 \end{cases}$
b. $t=1\ sec$, $(122.9,73.0)$.
$t=2\ sec$, $(245.7,111.1)$.
$t=3\ sec$, $(368.6,117.1)$.
c. $5.4\ sec$, $665.0\ ft$
d. see explanations.
Work Step by Step
Given $v_0=150\ ft/sec, \theta=35^\circ, h_0=3\ ft$, we have:
a. We can write the parametric equations for the position of the ball as
$\begin{cases} x=v_0\ cos\theta\ t\\ y=h_0+v_0\ sin\theta\ t -16t^2 \end{cases}$
where $t$ is the time and $-16\ ft/s^2$ is due to the gravity of the Earth. Thus we have
$\begin{cases} x=(150\ cos35^\circ) t\\ y=3+(150\ sin35^\circ) t -16t^2 \end{cases}$
b. (i) For $t=1\ sec$, we have
$\begin{cases} x=(150\ cos35^\circ) (1)\approx122.9\ ft \\ y=3+(150\ sin35^\circ) (1) -16(1)^2\approx73.0\ ft \end{cases}$
which gives point $(122.9,73.0)$.
(ii) For $t=2\ sec$, we have
$\begin{cases} x=(150\ cos35^\circ) (2)\approx245.7\ ft \\ y=3+(150\ sin35^\circ) (2) -16(2)^2\approx111.1\ ft \end{cases}$
which gives point $(245.7,111.1)$.
(iii) For $t=3\ sec$, we have
$\begin{cases} x=(150\ cos35^\circ) (3)\approx368.6\ ft \\ y=3+(150\ sin35^\circ) (3) -16(3)^2\approx117.1\ ft \end{cases}$
which gives point $(368.6,117.1)$.
c. To find the total flying time of the ball, let $y=0$. We have $3+(150\ sin35^\circ) t -16t^2=0$, which can be solved by either using the quadratic formula or graphically as shown. We have $t\approx5.4\ sec$ (discard the negative solution) and $x=(150\ cos35^\circ) (5.4)\approx665.0\ ft$ and this answer is consistent with the graph in the exercise.
d. From the graph in the exercise, it appears that the ball will reach a maximum height of about $120\ ft$ at a horizontal distance of about $330\ ft$. Algebraically, we can find the maximum in $y$ at $t=-\frac{150\ sin35^\circ}{2(-16)}\approx2.7\ sec$, which gives $y=3+(150\ sin35^\circ) (2.7) -16(2.7)^2\approx118.7\ ft$ and $x=(150\ cos35^\circ) (2.7)\approx331.8\ ft$, which agrees with the graph in the exercise.
