Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 73

Answer

$a = \frac{gh}{d}$

Work Step by Step

We can find the required speed at the top of the tube as; $v^2 = 0+2gh$ $v = \sqrt{2gh}$ We can find the required acceleration in the tube: $a = \frac{v^2-v_0^2}{2d}$ $a = \frac{(\sqrt{2gh})^2-0}{2d}$ $a = \frac{gh}{d}$
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