Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 50

Answer

(a) The car lands 61 meters from the base of the cliff. (b) The impact speed is 31 m/s

Work Step by Step

(a) We can find the vertical component of velocity when the car lands on the ground. Note that downward is the negative direction: $v_y^2 = v_{0y}^2+2ay$ $v_y = \sqrt{v_{0y}^2+2ay}$ $v_y = \sqrt{[(20~m/s)~sin(20^{\circ})]^2+(2)(-9.80~m/s^2)(-30~m)}$ $v_y = -25.2~m/s$ We can find the time it takes for the car to land on the ground. $t = \frac{v_y-v_{0y}}{g}$ $t = \frac{-25.2~m/s-(20~m/s)~sin(20^{\circ})}{-9.80~m/s^2}$ $t = 3.27~s$ We can find the horizontal distance the car travels in this time. $x = v_x~t$ $x = (20~m/s)~cos(20^{\circ})(3.27~s)$ $x = 61~m$ The car lands 61 meters from the base of the cliff. (b) We can find the car's impact speed. $v = \sqrt{(v_x)^2+(v_y)^2}$ $v = \sqrt{[(20~m/s)~cos(20^{\circ})]^2+(-25.2~m/s)^2}$ $v = 31~m/s$ The impact speed is 31 m/s.
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