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 - Problems - Page 45: 54

Answer

52 meters.

Work Step by Step

Let down be the positive direction and y = 0 be the initial height. The initial velocity is 0, and the acceleration is $g = 9.8\; m/s^2$. The final position is the unknown, y = H. The time of fall is $t_1$. Apply equation 2–11b. $$H=0+0+\frac{1}{2}gt_1^2$$ For the sound wave, we know that $v_{sound}=\frac{H}{3.4s-t_1}$. The time of travel for the sound wave is the total travel times of 3.4 seconds, minus the drop time. This can be re-arranged to solve for $t_1=3.4s-\frac{H}{v_{sound}}$. Substitute this expression into the earlier equation and solve for H. $$H=0+0+\frac{1}{2}g(3.4s-\frac{H}{v_{sound}})^2$$ This results in a quadratic equation for H. $$\frac{g}{2v_s^2}H^2-(\frac{g(3.4s)}{v_{sound}}+1)H+\frac{1}{2}g(3.4s)^2=0$$ Putting in numbers, we obtain H = 52m and H = 25900 m. The second solution is not correct because it gives a negative time for the rock to fall. The cliff is 52 meters high.
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