1. **State the problem:** Meg sees a dolphin jump out of the water with an initial velocity of 30 feet per second. We want to find how long the dolphin is in the air before it splashes back down. 2. **Formula:** The height of the dolphin at time $t$ seconds is given by the quadratic equation $$h = -16t^2 + vt + s$$ where: - $v$ is the initial velocity, - $s$ is the initial height (at water surface, $s=0$), - $h$ is the height at time $t$. 3. **Given values:** - $v = 30$ ft/s, - $s = 0$ ft (starting at water surface), - We want to find $t$ when $h=0$ (when dolphin hits water again). 4. **Set up the equation:** $$0 = -16t^2 + 30t + 0$$ 5. **Simplify:** $$-16t^2 + 30t = 0$$ 6. **Use quadratic formula:** The quadratic formula is $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = -16$, $b = 30$, and $c = 0$. 7. **Calculate discriminant:** $$b^2 - 4ac = 30^2 - 4(-16)(0) = 900 - 0 = 900$$ 8. **Calculate roots:** $$t = \frac{-30 \pm \sqrt{900}}{2(-16)} = \frac{-30 \pm 30}{-32}$$ 9. **Find each solution:** - For $+$ sign: $$t = \frac{-30 + 30}{-32} = \frac{0}{-32} = 0$$ (initial time) - For $-$ sign: $$t = \frac{-30 - 30}{-32} = \frac{-60}{-32} = \frac{60}{32} = \frac{15}{8} = 1.875$$ seconds 10. **Interpretation:** The dolphin is in the air from $t=0$ to $t=1.875$ seconds. 11. **Final answer:** To the nearest tenth, the dolphin is in the air for **1.9 seconds**.