Answer
(a) the position function of the ball $t$ seconds after it is hit is
${\bf{r}}\left( t \right) = \left( {10t + 85, - 5t + 20, - 16{t^2} + 25t} \right)$
(b) the ball hits the ground at position: $\simeq \left( {100.6,12.2,0} \right)$ from the center field.
From Figure 11 we see that the ball is still inside the field, however, it does not go in the goal before hitting the ground because the goal net's $x$-coordinate is located at $165$ ft from the center field.
Work Step by Step
(a) We have the initial velocity:
${\bf{v}}\left( 0 \right) = 10{\bf{i}} - 5{\bf{j}} + 25{\bf{k}}$
The center of the field is the origin of the $xy$-plane. The initial position is ${\bf{r}}\left( 0 \right) = \left( {85,20,0} \right)$.
The only force that acts on the ball is the gravitational force. So, the acceleration due to gravity is ${\bf{a}}\left( t \right) = - 32{\bf{k}}$ $ft/{s^2}$.
1. Find the velocity vector
We have
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {0,0, - 32} \right){\rm{d}}t = \left( {0,0, - 32t} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = 10{\bf{i}} - 5{\bf{j}} + 25{\bf{k}}$ gives
$\left( {10, - 5,25} \right) = \left( {0,0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {10, - 5,25} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {0,0, - 32t} \right) + \left( {10, - 5,25} \right)$
${\bf{v}}\left( t \right) = \left( {10, - 5, - 32t + 25} \right)$
2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {10, - 5, - 32t + 25} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {10t, - 5t, - 16{t^2} + 25t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {85,20,0} \right)$ gives
$\left( {85,20,0} \right) = \left( {0,0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {85,20,0} \right)$
Thus, the position function of the ball $t$ seconds after it is hit is
${\bf{r}}\left( t \right) = \left( {10t, - 5t, - 16{t^2} + 25t} \right) + \left( {85,20,0} \right)$
${\bf{r}}\left( t \right) = \left( {10t + 85, - 5t + 20, - 16{t^2} + 25t} \right)$
(b) From part (a) we obtain the position function of the ball $t$ seconds after it is hit:
${\bf{r}}\left( t \right) = \left( {10t + 85, - 5t + 20, - 16{t^2} + 25t} \right)$
The ball hits the ground if there exists $t$ such that
$ - 16{t^2} + 25t = 0$
$t\left( { - 16t + 25} \right) = 0$
The solutions are $t=0$ and $t = \frac{{25}}{{16}}$. The first solution corresponds to the time when the ball is kicked. So, we take the second solution $t = \frac{{25}}{{16}}$.
At $t = \frac{{25}}{{16}}$, the ball hits the ground at (from the center field)
${\bf{r}}\left( {\frac{{25}}{{16}}} \right) = \left( {\frac{{805}}{8},\frac{{195}}{{16}},0} \right) \simeq \left( {100.6,12.2,0} \right)$
From Figure 11 we see that the ball is still inside the field, however, it does not go in the goal before hitting the ground because the goal net's $x$-coordinate is located at $165$ ft from the center field.
