Chapter 9.3, Problem 39PS

Solution

$\left(a\right)$

To find: To use the formula given to complete the table.

Given information:

A sprinkler at ground level is used to water the garden. The water leaving the sprinkler has an initial speed of $25$ feet per second. Copy the table below. Use the given formula to complete the table.

  

Formula given is:

  $d=\frac{v^2}{32}\sin 2\theta$

Explanation:

The water leaving the sprinkler has an initial speed of $25$ feet per second.

  $v=25$

The horizontal distance $\left(d\right)$ water travelled is calculated as:

For $\theta ={25}^{\circ }$ :

  $\begin{array}{l}d=\frac{{25}^2}{32}\sin \left(2\times {25}^{\circ }\right)\\ =\frac{625}{32}\sin {50}^{\circ }\\ =\frac{625}{32}\times 0.7660\\ \approx 14.96\text{feet}\end{array}$

Similarly, we can determine the horizontal distance travelled by water for other value of $\theta$ :

$\theta$	${25}^{\circ }$	${30}^{\circ }$	${35}^{\circ }$	${40}^{\circ }$	${45}^{\circ }$	${50}^{\circ }$	${55}^{\circ }$	${60}^{\circ }$	${65}^{\circ }$
$d$	$14.96$	$16.91$	$18.35$	$19.23$	$19.53$	$19.23$	$18.35$	$16.91$	$14.96$

$\left(b\right)$

To interpret: The value of $\theta$ to maximize the horizontal distance travelled by the water.

  $\theta ={45}^{\circ }$

Given information:

Interpret what value of $\theta$ appear to maximize the horizontal distance travelled by water. Use the formula for horizontal distance travelled and unit circle to explain why the answer obtained is correct.

Explanation:

The horizontal distance $\left(d\right)$ water travelled is maximum at $\theta ={45}^{\circ }$ .

The horizontal distance travelled is $\text{19}\text{.53}\text{feet}$ .

As the value of $\sin 2\theta =\sin \left(2\times {45}^{\circ }\right)=\sin {90}^{\circ }$ .

The value of $\sin {90}^{\circ }$ is $1$ and also from the unit circle, the value of sine will never be more than $1$ .

Hence, at $\theta ={45}^{\circ }$ maximize the horizontal distance travelled by water.

$\left(c\right)$

To find: To compare the horizontal distance travelled by water.

Given information:

Compare the horizontal distance travelled by the water when $\theta ={\left(45-k\right)}^{\circ }$ with the distance $\theta ={\left(45+k\right)}^{\circ }$ .

Explanation:

Let the value of $k={15}^{\circ }$ .

Then, we have:

  $\begin{array}{l}\theta ={45}^{\circ }-{15}^{\circ }={30}^{\circ }\\ \theta ={45}^{\circ }+{15}^{\circ }={60}^{\circ }\end{array}$

The horizontal distance at ${30}^{\circ }$ and ${60}^{\circ }$ is $\text{16}\text{.91}\text{feet}$ .

Hence, the horizontal distance is same.