Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 1

Answer

The rocket should be launched when the horizontal distance is 38.9 meters.

Work Step by Step

We can find the acceleration of the rocket. $\sum F = ma$ $F_{thrust}-mg=ma$ $a = \frac{F_{thrust}-mg}{m}$ $a = \frac{15.0~N-(0.800~kg)(9.80~m/s^2)}{0.800~kg}$ $a = 8.95~m/s^2$ We can find the time it takes for the rocket to reach a height of 30 meters. $y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2y}{a}}$ $t = \sqrt{\frac{(2)(30~m)}{8.95~m/s^2}}$ $t = 2.59~s$ We can find the distance the target travels in this time. $x = v~t$ $x = (15~m/s)(2.59~s)$ $x = 38.9~m$ The rocket should be launched when the horizontal distance is 38.9 meters.
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.