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 - Page 70: 36


The boy and the balloon will reach the same vertical level at the same time and collide.

Work Step by Step

Choose x = 0 and y = 0 to be the point of launch with speed v and initial angle $\theta$, and upward to be the positive y direction. Assume the boy in the tree starts at x = d, y = h. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. Write the equations of motion for the balloon and for the boy. $$x_{balloon} = (v cos \theta)(t)$$ $$y_{balloon} = 0 + (v sin \theta)(t) – 0.5 g t^{2}$$ $$y_{boy} = h – 0.5 g t^{2}$$ Use the balloon's horizontal motion to find the time for the balloon to travel a distance d. $$t = \frac{d}{(v cos \theta)}$$ Use that time and find the balloon’s vertical position. Use the same time to find the boy’s vertical position. They are exactly the same, showing that the boy is splashed by the balloon.
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