Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set - Page 455: 65

Answer

$100 \text{ feet}$

Work Step by Step

Substituting $ g=32 $ and $ v=80 $ in the given equation, $ v=\sqrt{2gh} ,$ results to \begin{array}{l}\require{cancel} 80=\sqrt{2(32)h} \\ 80=\sqrt{64h} \\ 80=\sqrt{8^2h} \\ 80=8\sqrt{h} \\ \dfrac{80}{8}=\sqrt{h} \\ 10=\sqrt{h} .\end{array} Squaring both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} (10)^2=(\sqrt{h})^2 \\ h=100 .\end{array} Hence, the distance the object has fallen, $h,$ is $ 100 \text{ feet} .$
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