University Physics with Modern Physics (14th Edition)

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

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 93: 3.13

Answer

(a) The car should be traveling at a speed of 24.1 m/s (b) The speed of the car just before it lands is 31.0 m/s

Work Step by Step

(a) The car falls from a height of 21.3 m to a height of 1.8 m. The vertical distance is 21.3 - 1.8 m which is 19.5 m. We can find the time it takes to fall freely. $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(19.5~m)}{9.80~m/s^2}}$ $t = 1.99~s$ We can find the speed required to cross the 48.0-meter wide river in a time of 1.99 seconds. $v_x = \frac{x}{t} = \frac{48.0~m}{1.99~s} = 24.1~m/s$ The car should be traveling at a speed of 24.1 m/s. (b) $v_x = 24.1~m/s$ $v_y = \sqrt{2ay} = \sqrt{(2)(9.80~m/s^2)(19.5~m)}$ $v_y = 19.5~m/s$ We can find the speed of the car. $v = \sqrt{v_x^2+v_y^2}$ $v = \sqrt{(24.1~m/s)^2+(19.5~m/s)^2}$ $v = 31.0~m/s$ The speed of the car just before it lands is 31.0 m/s
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