Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 14 - Exercises and Problems - Page 272: 62

Answer

$2 \sqrt{\frac{L}{g}}$

Work Step by Step

In a previous problem, we found: $v = \sqrt{yg}$ Using this, we find: $ t = \int_0^L \frac{dy}{\sqrt{yg}}= 2 \sqrt{\frac{L}{g}}$
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