Chapter 10.2, Problem 33E

a.

To determine

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

Answer to Problem 33E

  $S\left(t\right)=〈56.78t,81t-16t^2〉$

Explanation of Solution

Given information:

A base-ball leaves the bat at an angle $55°$ from the horizontal travelling at 90 feet per seconds. A 20 foot boundary fence is 230 feet from the batter in a direct line.

Calculation:

The horizontal position, $\theta =55°,v=99\text{ ft/s,}g=0{\text{ft/s}}^2$ .

  $\begin{array}{c}x=u_{x}t+\frac{1}{2}g{t}^2\\ x=99\cos \left(55°\right)t\\ x=56.78t\end{array}$

The vertical position, $\theta =55°,v=99\text{ ft/s,}g=32{\text{ft/s}}^2$

  $\begin{array}{c}y=u_{x}t+\frac{1}{2}g{t}^2\\ y=99\sin \left(55°\right)t-\frac{1}{2}\times 32\times t^2\\ y=81t-16t^2\end{array}$

Thus, the position vector at time $t$ will be $S\left(t\right)=〈56.78t,81t-16t^2〉$

b.

To determine

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

Answer to Problem 33E

  $v\left(t\right)=〈56.78,81-32t〉$

Explanation of Solution

Given information:

A base-ball leaves the bat at an angle $55°$ from the horizontal travelling at 90 feet per seconds. A 20 foot boundary fence is 230 feet from the batter in a direct line.

Calculation:

The position vector at time $t$ will be $S\left(t\right)=〈56.78t,81t-16t^2〉$ .

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

  $\begin{array}{c}v\left(t\right)=s\text{'}\left(t\right)\\ =〈56.78,81-32t〉\end{array}$

Hence, the velocity vector is $〈56.78,81-32t〉$ .

c.

To determine

To check: The baseball hit the point $\left(230,20\right)$ .

Answer to Problem 33E

No

Explanation of Solution

Given information:

A base-ball leaves the bat at an angle $55°$ from the horizontal travelling at 90 feet per seconds. A 20 foot boundary fence is 230 feet from the batter in a direct line.

Calculation:

The position vector at time $t$ will be $S\left(t\right)=〈56.78t,81t-16t^2〉$ .

The position of ball is at $\left(230,20\right)$ .

So,

  $\begin{array}{cccc}56.78t=230& \text{and}& & 81t-16t^2=20\\ t\approx 4.1& \text{and}& & t\approx 0.3,4.8\end{array}$

Since the value of time is not same in both case, the baseball does not hit the point.

No, it is not correct.

d.

To determine

To check: The baseball hit the point $\left(230,20\right)$ .

Answer to Problem 33E

The ball hit the fence at 4.8 second.

The ball clear the home run in 4.1 seconds.

Explanation of Solution

Given information:

A base-ball leaves the bat at an angle $55°$ from the horizontal travelling at 90 feet per seconds. A 20 foot boundary fence is 230 feet from the batter in a direct line.

Calculation:

The position vector at time $t$ will be $S\left(t\right)=〈56.78t,81t-16t^2〉$ .

The position of ball is at $\left(230,20\right)$ .

So,

  $\begin{array}{cccc}56.78t=230& \text{and}& & 81t-16t^2=20\\ t\approx 4.1& \text{and}& & t\approx 0.3,4.8\end{array}$

Since the ball hit 20 feet, the time is 4.8 seconds.

Since the ball cross home 230 feet way, the time is 4.1 seconds.

Now draw the graph when baseball will hit the fence.

  

e.

To determine

To check: The baseball hit the point $\left(230,20\right)$ .

Answer to Problem 33E

  $92.17\text{ft/s}$

Explanation of Solution

Given information:

A base-ball leaves the bat at an angle $55°$ from the horizontal travelling at 90 feet per seconds. A 20 foot boundary fence is 230 feet from the batter in a direct line.

Calculation:

The velocity vector of ball is $〈56.78,81-32t〉$ .

Put $t=4.8$ into velocity vector and obtained,

  $\begin{array}{c}v\left(4.8\right)=〈56.78,81-32\left(4.8\right)〉\\ =〈56.78,-72.6〉\end{array}$

The magnitude of the speed is,

  $\begin{array}{c}\text{speed}=\sqrt{{56.78}^2+{\left(-72.6\right)}^2}\\ =92.17\text{ft/s}\end{array}$

Hence, the speed of ball is $92.17\text{ft/s}$ .