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: 32

Answer

97.4m/s

Work Step by Step

Let x= 0 and y = 0 directly below where the package is launched, and let upward be the positive y direction. Find the time of flight. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(69.4 m/s)(cos 0 ^{\circ})(t) = 425 m.$$ The time t is 6.124 s. Now use equation 2.11b to relate the projectile’s initial vertical velocity to the time. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The package starts at a height of y = 235 m, and the final vertical position is 0 m. The initial vertical velocity is found to be -8.37 m/s, meaning the package must be thrown downward slightly. To find the final speed, find the 2 components of the velocity at that time. The horizontal velocity remains constant, (69.4 m/s). The vertical velocity at that time is found by using equation 2.11a. The projectile’s initial vertical velocity is -8.37 m/s. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The vertical velocity at t = 6.124 s is -68.4 m/s. Use the Pythagorean theorem to find the overall speed of $97.4 m/s.$
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