College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 153: 85

Answer

When the package hits the ground, the horizontal distance between the package and the helicopter is $23~m$

Work Step by Step

We can find the time to fall $18~m$: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(18~m)}{9.80~m/s^2}}$ $t = 1.917~s$ We can find the horizontal distance between the package and the helicopter when the package hits the ground. Note that we should use the relative velocity between the package and the helicopter for $v_x$: $x = v_x~t = (12~m/s)(1.917~s) = 23~m$ When the package hits the ground, the horizontal distance between the package and the helicopter is $23~m$
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