1. **State the problem:** We want to find the time $t$ when the ball hits the ground. This means finding $t$ such that the height $h(t) = 0$. 2. **Write the height function:** The height above the ground after $t$ seconds is given by $$h(t) = -16t^2 + 40t$$ 3. **Set the height equal to zero to find when the ball hits the ground:** $$-16t^2 + 40t = 0$$ 4. **Factor the equation:** $$t(-16t + 40) = 0$$ 5. **Solve each factor for zero:** - From $t = 0$, the ball is on the ground at the initial time. - From $-16t + 40 = 0$, solve for $t$: $$-16t + 40 = 0$$ $$-16t = -40$$ $$\cancel{-16}t = \cancel{-40} \Rightarrow t = \frac{40}{16}$$ 6. **Simplify the fraction:** $$t = \frac{40}{16} = \frac{5 \times 8}{2 \times 8} = \frac{5}{2} = 2.5$$ 7. **Interpret the results:** The ball hits the ground at $t = 0$ (initial kick) and again at $t = 2.5$ seconds. **Final answer:** The ball hits the ground after **2.5 seconds**. **Answer choice:** (A) 2.5