Okay, I have noticed a few questions here that haven't been answered so here goes: (note, I am not an expert, but I did just complete my 'gravity' unit in physics last week, so I would consider myself... informed)
What would the gravity be, if we were like.. really near the earth's core? Every way. Any one atom exerts some (very little) force on every other atom. If you were in the center of the Earth, you would feel weightless because you would be pulled equally in every direction. That said, the pressure of everything else around you would easily crush you, so I don't suggest trying it (at least, not without a protective suit).
To actually answer the question: NO NO NO NO NO! Gravity is NOT 9.8226. The force of gravity (g) on Earth is -9.8226 (- because it is 'down' - towards the center). Gravity (G - the universal gravity constant) is (about) 6.674 x 10^-11 m^3/kgs^2 (about .00000000006674 cubic meters per kilogram seconds squared). The force you feel due to gravity (already defined as 'g' - see above) is equal to G(m1m2)/(r^2)
In other words gravity = G(mass_of_object_1 * mass_of_object_2)/(radius_Squared)
Therefore, if you are in the center of the Earth, you would be at a radius of 0, so the answer would be undefined (cannot divide by 0). If you were simply closer to the center (say, .01 meters from the center) you would feel more gravity:
This:
6.6 x 10^-11(m1m2)/(.01^2)
instead of:
6.6 x 10^-11(m1m2)/((6.3781 x 10^6)^2)
As you can imagine, dividing by a number less then 1/100000000 of normal would yield a significantly higher number.
Jumping = death/hurt? Due to the conservation of energy, you would hit the ground with exactly the same speed as you jumped with (disregarding air resistance/terminal velocity). It would hurt you no more to fall, then it did to jump in the first place.
(Sorry about yelling)