1. The problem is to express the height $h$ as a function of time $t$ given by the quadratic equation $$h = -2t^2 + 11t + 6.$$ 2. This is a quadratic function in standard form $h = at^2 + bt + c$ where $a = -2$, $b = 11$, and $c = 6$. 3. The formula for the vertex of a parabola $y = ax^2 + bx + c$ is at $t = -\frac{b}{2a}$. This gives the time at which the height is maximum or minimum. 4. Calculate the vertex time: $$t = -\frac{11}{2 \times -2} = -\frac{11}{-4} = \frac{11}{4} = 2.75.$$ 5. Substitute $t = 2.75$ back into the equation to find the maximum height: $$h = -2(2.75)^2 + 11(2.75) + 6 = -2(7.5625) + 30.25 + 6 = -15.125 + 30.25 + 6 = 21.125.$$ 6. Therefore, the maximum height reached is $21.125$ units at time $t = 2.75$ units. 7. The parabola opens downward because $a = -2 < 0$, indicating the height reaches a maximum at the vertex.