# 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

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.