1. **State the problem:** A ball is thrown upward from the top of a building. Its height above the ground after $t$ seconds is given by the function $$h(t) = -5t^2 + 20t + 30.$$ We need to find the time when the ball reaches its maximum height and the value of that maximum height. 2. **Formula and rules:** The height function is a quadratic function of the form $$h(t) = at^2 + bt + c$$ where $a = -5$, $b = 20$, and $c = 30$. Since $a < 0$, the parabola opens downward, so the vertex represents the maximum point. The time at which the maximum height occurs is given by the vertex formula: $$t = -\frac{b}{2a}$$ 3. **Calculate the time of maximum height:** $$t = -\frac{20}{2 \times (-5)} = -\frac{20}{-10} = 2$$ seconds. 4. **Calculate the maximum height:** Substitute $t=2$ into the height function: $$h(2) = -5(2)^2 + 20(2) + 30 = -5(4) + 40 + 30 = -20 + 40 + 30 = 50$$ meters. 5. **Answer:** The ball reaches its maximum height of 50 meters at 2 seconds after it is thrown.