University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises: 2.73

Answer

(a) The speed is 6.69 m/s. (b) It goes 4.48 meters above the ground. (c) The time it takes to fall from the maximum height down to the height of the athlete's head is 0.735 seconds.

Work Step by Step

(a) $v^2 = v_0^2 + 2ay = 0 + 2ay$ $v = \sqrt{2ay} = \sqrt{(2)(35.0~m/s^2)(0.640~m)}$ $v = 6.69~m/s$ The speed is 6.69 m/s. (b) We can let $v_0 = 6.69~m/s$ for this part of the question. $(y-y_0) = \frac{v^2-v_0^2}{2g} = \frac{0-(6.69~m/s)^2}{(2)(-9.80~m/s^2)}$ $y- y_0 = 2.28~m$ $y = 2.28~m + 2.20~m = 4.48~m$ It goes 4.48 meters above the ground. (c) We can find the time it takes for the shot to fall from the maximum height of 4.48 m down to a height of 1.83 m. $y = y_0 - \frac{1}{2}gt^2$ $1.83~m = 4.48~m - (4.9~m/s^2)t^2$ $t^2 = \frac{1.83~m - 4.48~m}{-4.9~m/s^2}$ $t = \sqrt{\frac{1.83~m - 4.48~m}{-4.90~m/s^2}}$ $t = 0.735~s$ The time it takes to fall from the maximum height down to the height of the athlete's head is 0.735 seconds.
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