## Physics (10th Edition)

Published by Wiley

# Chapter 2 - Kinematics in One Dimension - Problems: 49

1.08 s

#### Work Step by Step

Let up be the positive direction: $x=ut+\frac{1}{2}at^{2}$ $(-3.00)=(2.50)t+\frac{1}{2}(-9.8)t^{2}$ $4.9t^{2}-2.50t-3.00=0$ Using quadratic formula: $t=\frac{-(-2.50)±\sqrt {((-2.50)^{2}-4(4.9)(-3.00)}}{2(4.9)}$ $t\approx1.08$ or $t\approx-0.57$ $t\approx1.08\; s$ since $t\gt0$

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