To find the domain of the ball when it reaches its maximum height, we need to determine the time at which the volleyball reaches its peak height. The equation given for the height \( h \) of the volleyball is:
\[ h = -16t^2 + 33t + 6 \]
This is a quadratic equation of the form \( h = at^2 + bt + c \), where \( a = -16 \), \( b = 33 \), and \( c = 6 \).
The time \( t \) at which the maximum height occurs can be found using the formula for the vertex of a quadratic equation, which is given by:
\[ t = -\frac{b}{2a} \]
Plugging in the values for \( a \) and \( b \):
\[ t = -\frac{33}{2 \cdot -16} = \frac{33}{32} \]
This means that the maximum height occurs at \( t = \frac{33}{32} \) seconds.
To define the domain of the function while the volleyball is in the air, we also need to determine when the volleyball hits the ground, which occurs when \( h = 0 \). We can find the time when the ball hits the ground by solving the equation:
\[ -16t^2 + 33t + 6 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
We have:
- \( a = -16 \)
- \( b = 33 \)
- \( c = 6 \)
Calculating the discriminant:
\[ b^2 - 4ac = 33^2 - 4 \cdot (-16) \cdot 6 = 1089 + 384 = 1473 \]
Now applying the quadratic formula:
\[ t = \frac{-33 \pm \sqrt{1473}}{-32} \] Calculating \( \sqrt{1473} \approx 38.4 \):
Substituting back into the quadratic formula gives us:
\[ t = \frac{-33 + 38.4}{-32} \quad \text{and} \quad t = \frac{-33 - 38.4}{-32} \] Calculating these two values:
- First root: \[ t_1 = \frac{5.4}{-32} \text{ (not meaningful, as time can't be negative)} \]
- Second root: \[ t_2 = \frac{-71.4}{-32} \approx 2.23 \text{ seconds} \]
Thus, the volleyball will be in the air from \( t = 0 \) to \( t \approx 2.23 \) seconds.
In conclusion, the domain of the ball when it reaches its maximum height is:
\[ [0, \frac{33}{32}] \text{ seconds.} \]