Fundamentals of Physics Extended (10th Edition)

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

Chapter 2 - Motion Along a Straight Line - Problems - Page 38: 94

Answer

$1.2~s$ before it reaches the ground, the rock is $~~34.1~meters~~$ above the ground.

Work Step by Step

We can find the time it takes the rock to fall $60~m$: $y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2y}{a}}$ $t = \sqrt{\frac{(2)(60~m)}{9.8~m/s^2}}$ $t = 3.5~s$ Note that: $~~3.5~s-1.2~s = 2.3~s$ We can find the distance the rock falls in a time of $2.3~s$: $y = \frac{1}{2}at^2$ $y = \frac{1}{2}(9.8~m/s^2)(2.3~s)^2$ $y = 25.9$ We can find the height above the ground $1.2~s$ before the rock reaches the ground: $60~m - 25.9~m = 34.1~m$ $1.2~s$ before it reaches the ground, the rock is $~~34.1~meters~~$ above the ground.
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