Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 105: 18

Answer

$\textbf{(a)}$ $h=0.378\text{ m}$. $\textbf{(b)}$ It would decrease.

Work Step by Step

$\textbf{(a)}$ The horizontal component of the speed is constant so we can calculate the time required for traveling $l=0.500\text{ m}$ horizontally: $$l=v_h t\Rightarrow t=\frac{l}{v_h}.$$ Now, since there is no initial vertical component of the velocity, we have for the length of the descend: $$h=\frac{1}{2}gt^2=\frac{gl^2}{2v_h^2}=\frac{9.81\text{ m/s}^2\cdot 0.500^2(\text{m})^2}{2\cdot1.80^2(\text{m/s})^2}=0.378\text{ m}.$$ $\textbf{(b)}$ If the initial speed is increased, then the sparrow would travel the given horizontal distance for less time $t'$. Since the vertical distance flown is proportional to the time squared it follows that the distance of fall would decrease.
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