Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 43

Answer

$t=\dfrac{-v_{0}\pm\sqrt{v^{2}_{0}+2gh}}{g}$

Work Step by Step

$h=\dfrac{1}{2}gt^{2}+v_{0}t$; for $t$ Take $h$ to the right side of the equation to set it equal to $0$: $\dfrac{1}{2}gt^{2}+v_{0}t-h=0$ Multiply the whole equation by $2$, to avoid working with fractions: $gt^{2}+2v_{0}t-2h=0$ This is a quadratic equation. Let's solve it using the quadratic formula, which is: $t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ For this particular equation, $a=g$, $b=2v_{0}$ and $c=-2h$ Substitute: $t=\dfrac{-2v_{0}\pm\sqrt{(2v_{0})^{2}-4g(-2h)}}{2g}$ Simplify: $t=\dfrac{-2v_{0}\pm\sqrt{4v^{2}_{0}+8gh}}{2g}$ $t=\dfrac{-2v_{0}\pm\sqrt{4(v^{2}_{0}+2gh)}}{2g}$ $t=\dfrac{-2v_{0}\pm2\sqrt{v^{2}_{0}+2gh}}{2g}$ $t=\dfrac{-v_{0}\pm\sqrt{v^{2}_{0}+2gh}}{g}$
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