Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 3 - Exercises and Problems - Page 48: 59


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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