Physics: Principles with Applications (7th Edition)

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

Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 27

Answer

a. 30.8 m b. 5.02 s c. 136 m d. 28.9 m/s

Work Step by Step

Let x= 0 and y = 0 where the projectile is launched, and let upward be the positive y direction. The projectile’s initial vertical velocity is $(36.6 m/s)(sin 42.2 ^{\circ})$. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. a. The vertical velocity at the top is zero. Use equation 2.11c to find the maximum height of 30.8 m. b. For the entire flight, the displacement is zero. Use equation 2.11b to find the time of 5.02 s. The other root, t = 0s, corresponds to the start when the displacement is also zero. c. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(36.6 m/s)(cos 42.2 ^{\circ})(5.017 s) = 136 m.$$ d. Find the 2 components of the velocity at that time. The horizontal velocity remains constant, $(36.6 m/s)(cos 42.2 ^{\circ}) = 27.11 m/s$. The projectile’s initial vertical velocity is $(36.6 m/s)(sin 42.2 ^{\circ})$. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The vertical velocity at t = 1.5 s is 9.89 m/s, found by using equation 2.11a. Use the Pythagorean theorem to find the overall speed of 28.9 m/s.
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