Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 77


t $=$ $\frac{ v_{0}±\sqrt {(v_{0})^{2}+ 64(s_{0}-h)}}{32}$

Work Step by Step

We need to solve the quadratic equation $h$ $=$ $-$$16$$t^{2}$ $+$ $v_{0}$$t$ $+$ $s_{0}$. Shift the $h$ from Left Hand Side (LHS) to Right Hand Side (RHS), so that the equation becomes, $-$$16$$t^{2}$ $+$ $v_{0}$$t$ $+$ $s_{0}$ $-$$h$ $=$ $0$ We will use the quadratic formula to solve this equation. The quadratic formula is as follows, $\frac{ -b ±\sqrt {b^{2}- 4ac}}{2a}$ Here $a$$=$ $-$16 $b$ $=$ $v_{0}$ $c$ $=$ $s_{0}$$-$$h$ Putting the following in quadratic formula, we get $\frac{ -(v_{0}) ±\sqrt {(v_{0})^{2}- 4(-16)(s_{0} - h)}}{2(-16)}$ Solving the above, we get t $=$ $\frac{ v_{0}±\sqrt {(v_{0})^{2}+ 64(s_{0}-h)}}{32}$
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