Chapter 9 - Quadratic Functions and Equations - 9-1 Quadratic Graphs and Their Properties - Practice and Problem-Solving Exercises - Page 538: 26

$\approx$1.581 seconds

The question asks for when the orange hits the water. The variable, $h$, is the height of the orange. Therefore, we must solve for when $h$ is equal to 0. $ 0 = -16t^{2}+40$ $-40 = -16t^{2}$ $\frac{40}{16} = t^{2}$ $\frac{5}{2} = t^{2}$ $\sqrt \frac{5}{2} = t$ $t \approx 1.581$ seconds