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: 21

Answer

The ball is in the air for 3.2 seconds before it hits the ground.

Work Step by Step

$y = y_0+v_0~t+\frac{1}{2}at^2$ $0 = 2.0~m+(15~m/s)~t+\frac{1}{2}(-9.80~m/s^2)~t^2$ $(4.90~m/s^2)~t^2-(15~m/s)~t-2.0~m = 0$ We can then use the quadratic formula to find $t$: $t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{-(-15)\pm \sqrt{(-15)^2-(4)(4.90)(-2.0)}}{(2)(4.90)}$ $t = 3.2~s, -0.13~s$ Since the negative value is unphysical, the solution is $t = 3.2~s$. The ball is in the air for 3.2 seconds before it hits the ground.
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