This goes far beyond my limited knowledge regarding the physics of black holes, but my colleague told me this today:
I doubt that if I stand on the surface of a black hole - inside the Schwarzschild radius - and a stone falls on my head from outside the Schwarzschild radius, it cannot come out again. Then I just have to throw it back up with the energy it fell on my head so that it leaves the black hole. The stone can accept this energy because he fell on my head with it. Otherwise, the physical laws of gravity no longer apply. Just try to let a stone vibrate in a gravitational field whose generating mass has no expansion and idealizes does not collide with the vibrating body. It does not get stuck in the point x=0, but flies out again with infinite energy against an infinite acceleration. And this in clearly finite time.
If you have managed this calculation, then the question arises, where the energy should go, if the stone can fall in, but not fall out again? Energy is simply the time integral of the acceleration equation after multiplying both sides by the corresponding velocity equation. Since due to the existence of the acceleration equation also the velocity equation exists, the energy a body has in a force field is fixed for every movement. And this energy is on the way there, the same as on the way back, because it depends only on the place and not on the time. Even if time is deformed and the path is compressed, energy and place remain firmly connected. So the stone comes out of the black hole.
I wanted to share this with the community in order to get a clearer perspective to his statement. Are his assumptions correct? If so is there proof and maybe further literature?
(As far as I remember, the jets coming from super massive black holes do not contain anything from beyond the Schwarzschild radius.)