1. **State the problem:** We have a model rocket launched from a deck, and its height $A$ in meters after $t$ seconds is given by the quadratic function $$A = -5t^2 + 100t + 15.$$ We need to find: a) The height of the deck. b) The height of the rocket after 2 seconds. c) The maximum height reached by the rocket. d) The time taken to reach this maximum height. 2. **Formula and rules:** The quadratic function is in the form $$A = at^2 + bt + c$$ where $a = -5$, $b = 100$, and $c = 15$. - The height at $t=0$ is the constant term $c$, which represents the initial height (height of the deck). - To find the height at any time $t$, substitute $t$ into the function. - The maximum height of a parabola opening downward ($a<0$) occurs at $$t = -\frac{b}{2a}.$$ Substitute this $t$ back into the function to find the maximum height. 3. **Find the height of the deck (a):** At $t=0$, $$A = -5(0)^2 + 100(0) + 15 = 15.$$ So, the height of the deck is 15 meters. 4. **Find the height after 2 seconds (b):** Substitute $t=2$: $$A = -5(2)^2 + 100(2) + 15 = -5(4) + 200 + 15 = -20 + 200 + 15 = 195.$$ The rocket is 195 meters high after 2 seconds. 5. **Find the time to reach maximum height (d):** Use $$t = -\frac{b}{2a} = -\frac{100}{2 \times (-5)} = -\frac{100}{-10} = 10.$$ The rocket reaches maximum height at 10 seconds. 6. **Find the maximum height (c):** Substitute $t=10$ into the function: $$A = -5(10)^2 + 100(10) + 15 = -5(100) + 1000 + 15 = -500 + 1000 + 15 = 515.$$ The maximum height reached by the rocket is 515 meters. **Final answers:** - a) Height of the deck: 15 meters - b) Height after 2 seconds: 195 meters - c) Maximum height: 515 meters - d) Time to reach maximum height: 10 seconds