Answer
(a) the position function of the ball $t$ seconds after it is thrown is
${\bf{r}}\left( t \right) = \left( {40t,35t, - 16{t^2} + 32t + 5} \right)$
(b) the player is in bounds when he receives the ball.
Work Step by Step
(a) We have the initial velocity:
${\bf{v}}\left( 0 \right) = 40{\bf{i}} + 35{\bf{j}} + 32{\bf{k}}$
Let the center of the field be the origin of the $xy$-plane. Since the ball leaves his hand at a height of $5$ ft, the initial position is ${\bf{r}}\left( 0 \right) = \left( {0,0,5} \right)$.
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 ${\bf{v}}\left( 0 \right) = 40{\bf{i}} + 35{\bf{j}} + 32{\bf{k}}$ gives
$\left( {40,35,32} \right) = \left( {0,0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {40,35,32} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {0,0, - 32t} \right) + \left( {40,35,32} \right)$
${\bf{v}}\left( t \right) = \left( {40,35, - 32t + 32} \right)$
2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {40,35, - 32t + 32} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {40t,35t, - 16{t^2} + 32t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,0,5} \right)$ gives
$\left( {0,0,5} \right) = \left( {0,0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {0,0,5} \right)$
Thus, the position function of the ball $t$ seconds after it is thrown is
${\bf{r}}\left( t \right) = \left( {40t,35t, - 16{t^2} + 32t} \right) + \left( {0,0,5} \right)$
${\bf{r}}\left( t \right) = \left( {40t,35t, - 16{t^2} + 32t + 5} \right)$
(b) From part (a) we obtain the position function of the ball $t$ seconds after it is thrown:
${\bf{r}}\left( t \right) = \left( {40t,35t, - 16{t^2} + 32t + 5} \right)$
The ball is caught by a player $5$ ft above the ground. So, we have the equation for the $z$-component
$5 = - 16{t^2} + 32t + 5$
$ - 16{t^2} + 32t = 0$
$t\left( { - 16t + 32} \right) = 0$
We obtain the solutions: $t=0$ and $t=2$. Now, $t=0$ corresponds to the time when the ball is thrown. Thus, we take the second solution $t=2$.
At $t=2$, the position of the ball is ${\bf{r}}\left( 2 \right) = \left( {80,70,5} \right)$ from the center of the field.
Since the field is $150$ ft in width (y-coordinate) and $300$ ft in length (x-coordinate), ${\bf{r}}\left( 2 \right) = \left( {80,70,5} \right)$ is still inside of the field. Hence, the player is in bounds when he receives the ball.
