How fast can manned spacecraft slingshot from Black Holes? I've read of Black Holes launching particles at over 99% the speed of light. Could manned spacecraft use Black Holes to slingshot ourselves at this speed, or would the G forces kill us?
Intuitively, I worry the inertia of turning the curve would kill the crew.
 A: This won't work unless the black hole is orbiting something else.  It doesn't make a lot of difference to this whether we use a black hole or any other type of body - it's just gravity and conservation fo momentum and energy at work.
If you just had an isolated body and approached it, a slingshot won't happen.  All that happens is that your trajectory will change (and so will that of the other body - conservation of momentum etc.)  You don't gain any speed.
If the body you approach is orbiting another (as planets in the solar system orbit the Sun), then you can do a slingshot.  What is different is that you are, in essence, "dragged along" with the planet's orbital motion.  By choosing direction of approach properly you can get that boost in velocity (a slingshot, or gravity assist) because of the "dragging along" effect.
Now in your case you just have an isolated body (a black hole).  That doesn't get you a gravty assist.  You can use it to change course but not gain energy.  Energy you gain in falling towards it will be lost climbing away.
A: See StephenG's answer for why eccentric orbit doesn't let you gain any more energy leaving the mass than you had when you started approaching the mass.
You will probably get vaporized by other high-energy particles particles in your path, but let's ignore those since you asked about gravity problems only.
There won't be any any felt gravitational acceleration or felt centripetal acceleration. These are exactly balanced regardless of orbital eccentricity. An object in an eccentric orbit is still an object in free-fall, which is to say, it's weightless.
There will be tidal acceleration problems caused by the greater gravity closer to the black hole than slightly farther away, which (if you're standing with your head away from the black hole and your feet towards it) will cause your feet to try to pull away from your head.
There is some tidal acceleration that will be hazardous to our astronauts. Call that $\Delta a_{max}$. We could determine this experimentally using a centrifuge and some volunteers. My intuition is that it's probably about the same as maximum safe linear acceleration, so somewhere under 10 gravities.
Acceleration due to gravity from an object of mass M is
$a(r) = \frac{GM}{r^2\sqrt{1-\frac{R_s}{r}}}$
where $R_s = \frac{2GM}{c^2}$
so the tidal acceleration across two meters is
$\Delta a = a(r) - a(r+2m)$
And our distance of closest approach, r is a function of M obtained by solving for r in
$\Delta a_{max} = |\frac{GM}{r^2\sqrt{1-\frac{R_s}{r}}} - \frac{GM}{(r+2m)^2\sqrt{1-\frac{R_s}{(r+2m)}}}|$
I don't think this is solvable without a computer approximation. We can approximate by using the derivative,
$\Delta a \approx \Delta r \frac {d}{dr}a = -\dfrac{GM\left(4r-3R_s\right)\Delta r}{2\left(1-\frac{R_s}{r}\right)^\frac{3}{2}r^4} \approx -\dfrac{2GM\Delta r}{r^3}$
Which then becomes easy to solve,
$r_{min} \approx \sqrt[3]{|\frac{2GM\Delta r}{\Delta a_{max}}|}$
Checking to make sure that $r_{min} \gg R_s$ as a sanity-check after the fact.
We can then plug $r_{min}$ back into $a(r)$ to obtain our maximum safe rate of gravitational acceleration.
