College Physics (7th Edition)

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

Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 99: 44

Answer

$a). 0.64s, $ $b). 0.64m.$

Work Step by Step

a). Initial vertical component u=0. Thus: $S=ut+\frac{1}{2}at^{2}$ $2=0+0.5\times9.8t^{2}$ $t=0.64s$ b). Horizontal speed = 1 m/s. So distance traversed in 0.64s =$ 0.64\times 1=0.64m$
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