To find the volume of the pyramid given the information that it has the same height and base area as the cube, we start with the volume of the cube.
The formula for the volume of a cube is:
\[ V_{cube} = s^3 \]
where \( s \) is the length of a side of the cube. Since we know the volume of the cube is 9 cubic meters, we can find the side length:
\[ s^3 = 9 \]
Taking the cube root of both sides:
\[ s = \sqrt[3]{9} \]
Next, we know that the height \( h \) and base area \( A \) of the pyramid are equal to the corresponding measurements of the cube. The height of the cube is equal to its side length \( s \), which is \( \sqrt[3]{9} \).
The base area of the cube is \( s^2 \):
\[ A = s^2 = \left(\sqrt[3]{9}\right)^2 = \sqrt[3]{9^2} = \sqrt[3]{81} \]
The volume of the pyramid is calculated using the formula:
\[ V_{pyramid} = \frac{1}{3} \times A \times h \]
Substituting in \( A = \sqrt[3]{81} \) and \( h = \sqrt[3]{9} \):
\[ V_{pyramid} = \frac{1}{3} \times \sqrt[3]{81} \times \sqrt[3]{9} \]
Using the property of exponents, we can simplify this further. The product of the bases under cube roots can be combined:
\[ \sqrt[3]{81} \times \sqrt[3]{9} = \sqrt[3]{81 \times 9} = \sqrt[3]{729} \]
Calculating \( V_{pyramid} \):
\[ V_{pyramid} = \frac{1}{3} \times \sqrt[3]{729} \]
Since \( 729 = 9^3 \):
\[ V_{pyramid} = \frac{1}{3} \times 9 = 3 , \text{m}^3 \]
Thus, the volume of the pyramid is:
\[ \boxed{3 , \text{m}^3} \]
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