Chapter 10.2, Problem 34E

a.

To determine

To find: The position vector of the baseball at time $t$ .

Answer to Problem 34E

  $S\left(t\right)=〈44.12t,67.93t-16t^2〉$

Explanation of Solution

Given information:

A player punts the ball from his own 30-yard line at an angle of $57$ degree from horizontal, traveling at 81 feet per second. A 6-foot player from the opposing team is standing on his own 10-yard line on receive the punt.

Calculation:

The horizontal position, $\theta =57°,v=81\text{ ft/s,}g=0{\text{ft/s}}^2$ .

  $\begin{array}{c}x=u_{x}t+\frac{1}{2}g{t}^2\\ x=81\cos \left(57°\right)t\\ x=44.12t\end{array}$

The vertical position, $\theta =57°,v=81\text{ ft/s,}g=-32{\text{ft/s}}^2$

  $\begin{array}{c}y=u_{y}t+\frac{1}{2}g{t}^2\\ y=81\sin \left(57°\right)t-\frac{1}{2}\times 32\times t^2\\ y=67.93t-16t^2\end{array}$

Thus, the position vector at time $t$ will be $S\left(t\right)=〈44.12t,67.93t-16t^2〉$

b.

To determine

To find: The velocity vector of the baseball at time $t$ .

Answer to Problem 34E

  $v\left(t\right)=〈44.12,67.93-32t〉$

Explanation of Solution

Given information:

A player punts the ball from his own 30-yard line at an angle of $57$ degree from horizontal, traveling at 81 feet per second. A 6-foot player from the opposing team is standing on his own 10-yard line on receive the punt.

Calculation:

The position vector at time $t$ will be $S\left(t\right)=〈44.12t,67.93t-16t^2〉$ .

The velocity vector by derivative of position vector with respect to $t$ ,

  $\begin{array}{c}v\left(t\right)=s\text{'}\left(t\right)\\ =〈44.12,67.93-32t〉\end{array}$

Hence, the velocity vector is $〈44.12,67.93-32t〉$ .

c.

To determine

To find: The time when ball over the player on the line 10-yard line.

Answer to Problem 34E

0.53 second

Explanation of Solution

Given information:

A player punts the ball from his own 30-yard line at an angle of $57$ degree from horizontal, traveling at 81 feet per second. A 6-foot player from the opposing team is standing on his own 10-yard line on receive the punt.

Calculation:

The position vector at time $t$ will be $S\left(t\right)=〈44.12t,67.93t-16t^2〉$ .

The horizontal position of player 10-yard.

As 3 ft is 1 yard, So the 10-yard.

The number of feet in 10-yard is 30 feet.

So,

  $\begin{array}{c}30=56.78t\\ t=0.53\end{array}$

Thus, the ball over the player in 0.53 second.

d.

To determine

To check: The player catches the ball at $t=0.53\text{ second}$ .

Answer to Problem 34E

No

Explanation of Solution

Given information:

A player punts the ball from his own 30-yard line at an angle of $57$ degree from horizontal, traveling at 81 feet per second. A 6-foot player from the opposing team is standing on his own 10-yard line on receive the punt.

Calculation:

The position vector at time $t$ will be $S\left(t\right)=〈44.12t,67.93t-16t^2〉$ .

Put $t=0.53\text{ second}$ into $y=67.93t-16t^2$ .

  $\begin{array}{c}y=67.93\left(0.53\right)-16{\left(0.53\right)}^2\\ =31.51\text{ ft}\end{array}$

Since the height of player is 6 ft, the ball height is at 31.51 ft.

So, the player cannot catch the ball.