## Physics (10th Edition)

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 $t\gt0$