Why does the jet come out perpendicular to the accretion disc? Why does the jet always come out perpendicular to the plane of the accretion disc around a black hole?
 A: This is a complicated question as it involves plasma physics. But the concept you're looking for is magnetocentrifugal acceleration
and concepts like Alfven surface etc. are involved.
What we will be looking at is the centrifugal-gravitational potential
\begin{equation}
\Phi(r,z) = - \frac{GM}{\sqrt{r^2+z^2}} - \frac{GM}{2r_0^3}r^2
\end{equation}
where $r,z$ can be thought of as like cylindrical coordinates.
These coordinates are in reference to a system moving with the magnetic surface and $r_0$ is the base point where the accretion disk intersects the magnetic surface. Hence the second term in the potential is due to comoving with the surface: $\frac{1}{2} \Omega^2r^2$ (at supersonic speeds).
A magnetic surface is a surface of revolution due to the rotation of the poloidal magnetic field lines around the $z$-axis along which the magnetic flux is constant.

The picture, taken from Prof. S. De Rijcke's course on "The Physics of Galaxies", shows the centrifugal-gravitational potential. The base point is a saddle point of this surface: all points to the right of the equipotential surface $(r_0,0)$ are points where the fluid elements will be slung away.
The equation for the equipotential surface through the base point in units $GM/r_0$ and $r$ in units $r_0$ is given by
\begin{equation}
\frac{1}{\sqrt{r^2+z^2}}+\frac{r^2}{2} = \frac{3}{2}
\end{equation}
For $r \approx 1$ close to the base point we have that $z = \pm \sqrt{3}(r-1)$. Hence the equipotential surface makes an angle of 30 degrees with the vertical. If the magnetic surface has a larger angle then every element will be accelerated along the magnetic field lines. Since these field lines along with matter are confined to the magnetic surface this results in a collimated jet.
