To find the vertex of the quadratic function \( h(t) = -6t^2 + 33t + 6 \), we can use the formula for the vertex of a quadratic function in the form \( h(t) = at^2 + bt + c \), where the vertex \( t \) coordinate is given by:
\[ t = -\frac{b}{2a} \]
Here, \( a = -6 \) and \( b = 33 \). Plugging these values into the formula:
\[ t = -\frac{33}{2(-6)} = \frac{33}{12} = 2.75 \]
Now, we need to find the height \( h \) at \( t = 2.75 \). Plugging \( t = 2.75 \) back into the function:
\[ h(2.75) = -6(2.75^2) + 33(2.75) + 6 \]
Calculating \( 2.75^2 \):
\[ 2.75^2 = 7.5625 \]
Now substitute this back in:
\[ h(2.75) = -6(7.5625) + 33(2.75) + 6 \] \[ = -45.375 + 90.75 + 6 \] \[ = -45.375 + 96.75 = 51.375 \]
Thus, the vertex of the parabola, which gives the maximum height of the volleyball, is \((2.75, 51.375)\).
So, the correct ordered pair that represents the vertex is:
\[ \boxed{(2.75, 51.375)} \]