Question 2 In the tetrahedron $P-ABC$, the base is an equilateral triangle with side length 3, $PA=3, PB=4, PC=5$. Then the volume $V$ of the tetrahedron $P-ABC$ is $\qquad$
【Analysis】A tetrahedron has four faces, and each face can serve as the base, so the method is not unique.

Solution As shown in Figure 1, it is easy to know that $\triangle P B C$ is a right triangle, and we take the midpoint $O$ of the hypotenuse $P C$.

Since $A P=A B=A C$, point $A$'s projection on plane $P B C$ is the circumcenter $O$ of the right triangle $\triangle P B C$. Connect $A O$. Then
$A O \perp$ plane $P B C$.
Therefore, $V=\frac{1}{3} S_{\triangle P B C} \cdot A O$
$=\frac{1}{3}\left(\frac{1}{2} \times 3 \times 4\right) \sqrt{3^{2}-\left(\frac{5}{2}\right)^{2}}=\sqrt{11}$.