Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Exercises and Problems - Page 87: 29


$137.03 \ m/s$

Work Step by Step

We know that the y-forces cancel out, for the height of the plane is constant throughout the bank. Thus, we find: $mg = F_n cos\theta$ We know that the x-component of the normal force due to the air is the centripetal force, so we find: $F_nsin\theta=\frac{mv^2}{r}$ Combining these equations gives: $mg = \frac{mv^2}{rtan\theta}$ This means that the velocity is: $v=\sqrt{grtan\theta}=\sqrt{(9.81)(3600)(tan28)}=137.03 \ m/s$
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