Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 2 - Exercises and Problems: 80

Answer

Please see the work below.

Work Step by Step

We know that the length of the window is given as $d=vt-\frac{1}{2}gt^2$ We plug in the known values to obtain: $-1.3=v(0.22)-\frac{1}{2}(9.8)(0.22)^2$ $v=-4.831\frac{m}{s}$ Hence the height above the window is: $h=\frac{v^2}{2g}$ We plug in the known values to obtain: $h=\frac{(4.831)^2}{2(9.8)}$ $h=1.2m$
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