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 3 - Exercises and Problems - Page 51: 65


Please see the work below.

Work Step by Step

We know that $t=\sqrt{\frac{2h}{g}}$ $t=\sqrt{\frac{2(1.6-0.93)}{9.8}}=0.3698s$ Now we can find the horizontal speed of the water as $v=\frac{x}{t}$ We plug in the known values to obtain: $v=\frac{2.1}{0.3698}=5.7\frac{m}{s}$
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