Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.5 Further Applications of Right Triangles - 2.5 Exercises - Page 85: 44b

Answer

As $v$ increases from $43~ft/s$ to $45~ft/s$, the distance increases from $64.4 ~ft$ to $69.9 ~ft$. As $v$ increases, the distance increases significantly.

Work Step by Step

$D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2+64~h}}{32}$ We can find $D$ when $v = 43~ft/s$: $D = \frac{(43~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+43~ft/s~cos~42^{\circ}~\sqrt{(43~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$ $D = 64.4~ft$ We can find $D$ when $v = 44~ft/s$: $D = \frac{(44~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+44~ft/s~cos~42^{\circ}~\sqrt{(44~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$ $D = 67.1~ft$ We can find $D$ when $v = 45~ft/s$: $D = \frac{(45~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+45~ft/s~cos~42^{\circ}~\sqrt{(45~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$ $D = 69.9~ft$ As $v$ increases from $43~ft/s$ to $45~ft/s$, the distance increases from $64.4 ~ft$ to $69.9 ~ft$. As $v$ increases, the distance increases significantly.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.