Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 93: 132a

Answer

$v_0 = (10~m/s)\hat{i}+(10~m/s)\hat{j}$

Work Step by Step

On the graph, we can see that the shot moves a horizontal distance of $10~m$ in a time of $1.0~s$. Thus $v_x = 10~m/s$ On the graph, we can see that the shot reaches maximum height in a time of $1.25~s$ We can find an expression for $g$: $1.25~s = \frac{v_{0y}}{g}$ $g = \frac{v_{0y}}{1.25~s}$ The shot moves up a vertical distance of $4.0~m$ in a time of $0.50~s$. We can find $v_{0y}$: $y = y_0+v_{0y}~t-\frac{1}{2}gt^2$ $y-y_0 = v_{0y}~t-\frac{1}{2}(\frac{v_{0y}}{1.25~s})(t)^2$ $4.0~m = v_{0y}~(t-\frac{t^2}{2.5~s})$ $v_{0y} = \frac{4.0~m}{t-\frac{t^2}{2.5~s}}$ $v_{0y} = \frac{4.0~m}{0.50~s-\frac{(0.50~s)^2}{2.5~s}}$ $v_{0y} = 10~m/s$ We can express the initial velocity in unit-vector notation: $v_0 = (10~m/s)\hat{i}+(10~m/s)\hat{j}$
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