College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 64: 70

Answer

The water balloon will NOT hit her. It will come close by $1-0.819=0.181m$ at the height level of her head.

Work Step by Step

Height for the water balloon to travel $= 18-1.7=16.3m$ $S=ut+\frac{1}{2}gt^{2}$ So, $16.3=0+0.5\times 9.8\times t^{2}$ $t=1.82s$ So, in 1.82s, the professor will move $=1.82\times 0.45=0.819m$ which is less than 1m. So, the water balloon will NOT hit her. It will come close by $1-0.819=0.181m$ at the height level of her head.
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