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 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 38

Answer

(a) $v = \sqrt{(2h)(\frac{F_{thrust} - mg}{m})}$ (b) $v = 54.3~m/s$

Work Step by Step

(a) We can find an expression for the acceleration of the rocket. $\sum F = ma$ $F_{thrust} - mg = ma$ $a = \frac{F_{thrust} - mg}{m}$ We can find the speed when the rocket reaches a height $h$. $v^2 = 2ah$ $v = \sqrt{2ah}$ $v = \sqrt{(2h)(\frac{F_{thrust} - mg}{m})}$ (b) We can use the expression in part (a) to find the speed of the rocket. $v = \sqrt{(2h)(\frac{F_{thrust} - mg}{m})}$ $v = \sqrt{(2)(85~m)[\frac{9.5~N - (0.350~kg)(9.80~m/s^2}{0.350~kg}]}$ $v = 54.3~m/s$
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.