To find the vertex of the quadratic function \( h(t) = -6t^2 + 33t + 6 \), we can use the vertex formula for a quadratic function \( h(t) = at^2 + bt + c \), where the x-coordinate of the vertex is given by:
\[ t = -\frac{b}{2a} \]
In this case, \( a = -6 \) and \( b = 33 \). Plugging in these values:
\[ t = -\frac{33}{2 \cdot -6} = \frac{33}{12} = 2.75 \]
Next, we calculate the height \( h \) at \( t = 2.75 \) by substituting it back into the height function:
\[ h(2.75) = -6(2.75)^2 + 33(2.75) + 6 \]
Calculating \( (2.75)^2 = 7.5625 \):
\[ h(2.75) = -6(7.5625) + 33(2.75) + 6 \] \[ = -45.375 + 90.75 + 6 \] \[ = 51.375 \]
Thus, the vertex of the function is at the point \( (2.75, 51.375) \).
The correct ordered pair representing the vertex is:
\[ \boxed{(2.75, 51.375)} \]