Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 46: 68


The two stones are 23.8 meters apart.

Work Step by Step

Let's assume that the down direction is positive. When the speed of the second stone is 12.0 m/s, we can find the distance $y_2$ the second stone has dropped: $y_2 = \frac{v^2 - v_0^2}{2g} = \frac{(12.0 ~m/s)^2 - 0}{(2)(9.80 ~m/s^2)} = 7.35 ~m$ We can then find the time it takes the second stone to reach a speed of 12.0 m/s: $t = \frac{12.0 ~m/s}{9.80 ~m/s^2} = 1.22 ~s$ When the second stone has been falling for 1.22 seconds, the first stone has been falling for 2.52 seconds. We can find the distance $y_1$ that the first stone has dropped when t = 2.52 seconds: $y_1 = \frac{1}{2}at^2 = \frac{1}{2}(9.80 ~m/s^2)(2.52 ~s)^2 = 31.1~m$ $\Delta y = y_1 - y_2 = 31.1 ~m - 7.35 ~m = 23.8 ~m$ The two stones are 23.8 meters apart.
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