A volleyball is served by a 6-foot player at an initial upward velocity of 33 feet per second. The situation is modeled by the equation h=−16t2+33t+6

To find the time when the volleyball reaches its maximum height, we can use the formula for the vertex of a quadratic equation in the form \(h(t) = at^2 + bt + c\), where the maximum or minimum value occurs at \(t = -\frac{b}{2a}\).

In this case, the given equation is: \[ h(t) = -16t^2 + 33t + 6 \]

Here, \(a = -16\) and \(b = 33\). Plugging these values into the formula for \(t\):

\[ t = -\frac{b}{2a} = -\frac{33}{2(-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]

We can round this to about 1.03 seconds.

Now, the domain for the time \(t\) when the ball reaches maximum height starts at \(t = 0\) (when the ball is served) and goes up to approximately \(t = 1.03\) seconds (when it reaches maximum height).

The domain of the situation when the ball reaches its maximum height is from \(t = 0\) to \(t = 1.03\) seconds.

Thus, the answer would be:

1.03 seconds.