Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.9 - Propulsion with Variable Mass - Problems - Page 312: 146



Work Step by Step

The required maximum speed can be determined as follows: $t=\frac{m_f}{q}=\frac{300}{15}=20s$ Now, $v_{max}=v_e\ln(\frac{W_r+W_f}{W_r+W_r-qt})$ We plug in the known values to obtain: $v_{max}=4400\ln (\frac{500+300}{500+300-15(20)})$ This simplifies to: $v_{max}=2068ft/s$
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.