Imagine we have something very heavy (i.e supermassive black hole) and some object that we can throw with 0.999999 speed of light (i.e proton). We are throwing our particle in the direction of hole. The black hole is so heavy that we can assume that in some moment acceleration of gravity would be say 0.0001 speed of light/s^2. So the question is what will be the speed of proton in few a seconds later, assuming we have such distances that it will not hit the black hole before.
Answer
This the classic "hurling a stone into a black hole" problem. It's described in detail in sample problem 3 in chapter 3 of Exploring Black Holes by Edwin F.Taylor and John Archibald Wheeler. Incidentally I strongly recommend this book if you're interested in learning about black holes. It does require some maths, so it's not a book for the general public, but the maths is fairly basic compared to the usual GR textbooks.
The answer to your question is that no-one observes the stone (proton in your example) to move faster than light, no matter how fast you throw it towards the black hole.
I've phrased this carefully because in GR it doesn't make sense to ask questions like "how fast is the stone" moving unless you specify what observer you're talking about. Generally we consider two different types of observer. The Schwarzschild observer sits at infinity (or far enough away to be effectively at infinity) and the shell observer sits at a fixed distance from the event horizon (firing the rockets of his spaceship to stay in place).
These two observers see very different things. For the Schwarzschild observer the stone initially accelerates, but then slows to a stop as it meets the horizon. The Schwarzschild observer will never see the stone cross the event horizon, or not unless they're prepared to wait an infinite time.
The shell observer sees the stone fly past at a velocity less than the speed of light, and the nearer the shell observer gets to the event horizon the faster they see the stone pass. If the shell observer could sit at the event horizon (they can't without an infinitely powerful rocket) they'd see the stone pass at the speed of light.
To calculate the trajectory of a hurled stone you start by calculating the trajectory of a stone falling from rest at infinity. I'm not going to repeat all the details from the Taylor and Wheeler book since they're a bit involved and you can check the book. Instead I'll simply quote the result:
For the Schwarzschild observer:
$$ \frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \left( \frac{2M}{r} \right)^{1/2} $$
For the shell observer:
$$ \frac{dr_{shell}}{dt_{shell}} = - \left( \frac{2M}{r} \right)^{1/2} $$
These equations use geometric units so the speed of light is 1. If you put $r = 2M$ to find the velocities at the event horizon you'll find the Schwarzschild observer gets $v = 0$ and the (hypothetical) shell observer gets $v = 1$ (i.e. $c$).
But this was for a stone that started at rest from infinity. Suppose we give the stone some extra energy by throwing it. This means it corresponds to an object that starts from infinity with a finite velocity $v_\infty$. We'll define $\gamma_\infty$ as the corresponding value of the Lorentz factor. Again I'm only going to give the result, which is:
For the Schwarzschild observer:
$$ \frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \left[ 1 - \frac{1}{\gamma_\infty^2}\left( 1 - \frac{2M}{r} \right) \right]^{1/2} $$
For the shell observer:
$$ \frac{dr_{shell}}{dt_{shell}} = - \left[ 1 - \frac{1}{\gamma_\infty^2}\left( 1 - \frac{2M}{r} \right) \right] ^{1/2} $$
Maybe it's not obvious from a quick glance at the equations that neither $dr/dt$ nor $dr_{shell}/dt_{shell}$ exceeds infinity, but if you increase your stone's initial velocity to near $c$ the value of $\gamma_\infty$ goes to $\infty$ and hence 1/$\gamma^2$ goes to zero. In this limit it's easy to see that the velocity never exceeds $c$.
In his comments Jerry says several times that the velocity exceeds $c$ only after crossing the event horizon. While Jerry knows vaaaaastly more than me about GR I would take him to task for this. It certainly isn't true for the Schwarzschild observer, and you can't even in principle have a shell observer within the event horizon.
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