Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 112: 113a


time taken $\approx$ 5.86 seconds

Work Step by Step

$Change in height = H-h = ut + \frac{1}{2}at^{2}$ Initial velocity (u) for free fall=0 Final height (H) is 0 Initial height (h) is 550 The time taken to reach the ground is given as t Plug in values $0= 550 + 0*t -16t^{2}$ $550 =16t^{2}$ $t^{2}=550/16$ $t=\sqrt {550/16}$$\approx$ 5.86 seconds
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.