Answer
a. $\begin{cases} x=(180\ cos40^\circ) t\\ y=3+(180\ sin40^\circ) t -16t^2 \end{cases}$
b. $t=1\ sec$, $(137.9,102,7)$.
$t=2\ sec$, $(275.8,170.4)$.
$t=3\ sec$, $(413.7,206.1)$.
c. $ 7.3\ sec$, $1006.6\ ft$
d. see explanations.
Work Step by Step
Given $v_0=180\ ft/sec, \theta=40^\circ, h_0=3\ ft$, 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
$\begin{cases} x=(180\ cos40^\circ) t\\ y=3+(180\ sin40^\circ) t -16t^2 \end{cases}$
b. (i) For $t=1\ sec$, we have $\begin{cases} x=(180\ cos40^\circ) (1)\approx137.9\ ft \\ y=3+(180\ sin40^\circ) (1) -16(1)^2\approx102.7\ ft \end{cases}$
which gives the point $(137.9,102,7)$.
(ii) For $t=2\ sec$, we have
$\begin{cases} x=(180\ cos40^\circ) (2)\approx275.8\ ft \\ y=3+(180\ sin40^\circ) (2) -16(2)^2\approx170.4\ ft \end{cases}$
which gives the point $(275.8,170.4)$.
(iii) For $t=3\ sec$, we have
$\begin{cases} x=(180\ cos40^\circ) (3)\approx413.7\ ft \\ y=3+(180\ sin40^\circ) (3) -16(3)^2\approx206.1\ ft \end{cases}$
which gives point $(413.7,206.1)$.
c. To find the total flying time of the ball, let $y=0$. We have $3+(180\ sin40^\circ) t -16t^2=0$, which can be solved by either using the quadratic formula or graphically as shown. We have $t\approx7.3\ sec$ (discard the negative solution) and $x=(180\ cos40^\circ) (7.3)\approx1006.6\ ft$ 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 $200\ ft$ at a horizontal distance of about $500\ ft$. Algebraically, we can find the maximum in $y$ at $t=-\frac{180\ sin40^\circ}{2(-16)}\approx3.6\ sec$, which gives $y=3+(180\ sin40^\circ) (3.6) -16(3.6)^2\approx212.2\ ft$ and $x=(180\ cos40^\circ) (3.6)\approx496.4\ ft$, which agrees with the graph in the exercise.
