Chapter 8.7, Problem 8CFU

a.

To determine

To find: the parametric equations that represent the path of probes.

Answer to Problem 8CFU

  $\begin{array}{c}x=440t\\ y=-16t^2+3500\end{array}$

Explanation of Solution

Given information:

Height $\left(h\right)=3500\text{ feet}$

Initial velocity $\left(\overrightarrow{v}\right)=330\text{ mph}$

Path of plane is parallel to the ground at the time of releasing the probes, so $\theta =0°$ .

Calculation:

First convert the velocity into feet per second as follows:

  $\begin{array}{c}\left|\overrightarrow{v}\right|=330\text{ mph}\left(\frac{5280\text{ feet}}{1\text{ mile}}\right)\left(\frac{1\text{ hour}}{3600\text{ s}}\right)\\ =440\text{ feet per second}\end{array}$

Now solve the horizontal and vertical components as shown below.

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =t\left(440\right)\cos 0°\left[\text{Put the value of }\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ =440t\end{array}$

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2+h\\ =t\left(440\right)\sin 0°-\frac{1}{2}\left(32\right)t^2+3500\left[\text{Put the value of }\left|\overrightarrow{v}\right|,g,h\text{ and }\theta \right]\\ =-16t^2+3500\end{array}$

Thus the parametric equations are $x=440t$ and $y=-16t^2+3500$ .

b.

To determine

To graph: the parametric equations that represent the path of probes.

Explanation of Solution

Given information:

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{c}x=440t\\ y=-16t^2+3500\end{array}$

Calculation:

Express $t$ in terms of $x$ as shown below.

  $\begin{array}{l}x=440t\\ t=\frac{x}{400}\end{array}$

Now substitute the value of $t$ in parametric equation for $y$ as follows:

  $\begin{array}{c}y=-16t^2+3500\\ =-16{\left(\frac{x}{400}\right)}^2+3500\end{array}$

Create a table of values for $x\text{ and }y$ as shown below.

$x$	0	1000	2000	3000	4000	5000	6000	6507
$y$	3500	3417.3	3169.4	2756.2	2177.7	1433.8	524.8	0.7

Plot the points and join them with a soft curve to get the graph of given parametric equations as shown below.

  

c.

To determine

To find: the time taken by the probes to reach the ground.

Answer to Problem 8CFU

14.8 seconds

Explanation of Solution

Given information:

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{c}x=440t\\ y=-16t^2+3500\end{array}$

Calculation:

When the probe will reach the ground the value of $y$ will become zero. Substitute $y=0$ in the equation for $y$ and solve $t$ as shown below.

  $\begin{array}{c}y=-16t^2+3500\\ 0=-16t^2+3500\left[\text{Put }y=0\right]\\ 16t^2=3500\left[\text{Add 16}{t}^2\text{ on each side}\right]\\ t^2=\frac{3500}{16}\left[\text{Divide each side by 16}\right]\\ t=\sqrt{\frac{3500}{16}}\left[\text{Taking square roots on each side}\right]\\ =14.8\text{ seconds}\end{array}$

Thus the required time taken by probe to reach the ground is 14.8 seconds.

d.

To determine

To find: the horizontal distance travelled by probes before hitting the ground.

Answer to Problem 8CFU

  $6214\text{ feet}$

Explanation of Solution

Given information:

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{c}x=440t\\ y=-16t^2+3500\end{array}$

Time taken to reach the ground $\left(t\right)=14.8\text{ seconds}$

Calculation:

Substitute the value of $t$ in parametric equation for $x$ and solve as follows:

  $\begin{array}{c}x=440t\\ =440\left(14.8\right)\\ =6214\text{ feet}\end{array}$

Thus the horizontal distance travelled by probes before hitting the ground is 6215 feet.