Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - Chapter Review - Page 321: 41

Answer

See below

Work Step by Step

Given: $h_0=9\\v_0=-40\\h=0$ The formula for the height: $h=h_0+v_0t-16t^2$ Substitute: $-16t^2-40t-9=h$ Setting $h=0$, we have: $-16t^2-40t-9=0$ The solution is: $t=\frac{40 \pm \sqrt (-40)^2-4(-16)(9)}{2(-16)}\\=\frac{40 \pm \sqrt 2176}{-32}\\=\frac{-40 \pm \sqrt 2176}{-32}$ Hence, $t \approx 0.21 \lor t\approx-2.71$ Since time cannot be negative, we will choose $t=0.21$ sec
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