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: 35

Answer

No; Must be 34.7m-4.6m.

Work Step by Step

Choose x = 0 and y = 0 to be the point of launch and upward to be the positive y direction. The initial speed is 20 m/s. We can find both initial velocity components from the given information. x-component: $$(20 m/s)(cos 37 ^{\circ}) = 15.97 m/s$$ y-component: $$(20 m/s)(sin 37 ^{\circ}) = 12.04 m/s$$ Find the time to reach the goalposts. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(20 m/s)(cos 37 ^{\circ})(t) = 36.0 m$$ The time t is 2.254 s. Use this time, with the vertical information, in equation 2.11b to find the vertical position of the football when it reaches the location of the goalposts. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The final height is 2.24 m, so it does not pass over the bar. To find the horizontal positions from which a field goal can be kicked, start with the vertical motion and find the times at which the ball is at y = 3.05 m. Use equation 2.11b with the known initial vertical velocity and acceleration to find the 2 possible times of 2.170 s, 0.287 s. These correspond to the ball on the way up, and on the way down. With those times, return to the horizontal motion, and find the horizontal distance traveled during each of those times, at constant horizontal velocity. The distances are 4.6 m and 34.7 m. The kick may be made from anywhere within these distances, because the ball will be higher than 3.05 m.
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