[Physics] How does one exert greater force on the ground by jumping

Question

When one jumps, how does he/she manage to exert greater force on their ground than their weight?
Also, what is normal force and the reaction force (are they the same thing?) and by newton's third law, shouldn't the reaction(weight) when we are standing on the ground that the ground exerts on us send us flying above the ground- why doesn't the law apply here?
Finally, when we drop a hard stone on the ground why doesn't it bounce? Plus, why is the force exerted by the stone on the ground greater than its weight?

Answer 1

There's still something missing from all the answers so far. When you drop something on the ground, say, a rock of mass $m$, by the time it makes contact with the ground it's traveling at a velocity $v$ and thus has momentum $p = mv$. To be stopped completely, its momentum has to equal $0$ at the end. So you have a total change in momentum of $\Delta p$. According to (the most literal, I think) Newton's 2nd law, you have $\Delta p = F \Delta t$, where $F$ is the force slowing down the object over the timespan $\Delta t$ (in reality time is continuous and $F$ is probably changing continuously, but this is enough to illustrate the point).

So, if the $m = 1\ kg$ rock goes from falling at $v = 10\ m/s$ to $0$ in a millisecond or so, you might have $F = \Delta p/\Delta t = 10\ kg\ m/s /(.001s)=10000\ N$, which is obviously much bigger than just the gravitational force of $F_g \approx 1\ kg \times 10\ m/s^2 = 10\ N$.