The other answers explain things in terms of the activation energy barrier, but they don't explain why the activation energy barrier is lowest for K. I will attempt to do that here.
When a metal reacts with water, several things must happen (not necessarily in this order):
- One or more electrons have to be removed from the metal
- One or more $O-H$ bonds have to break on water molecules (one per electron)
- The electron(s) must be "picked up" by the $OH$ to become $OH^-$
As @ron said, the "vigorousness" of this process depends on the kinetics of the reaction, which are in turn determined by the activation energy of the rate-limiting step.
In the links that he provided (1,2), they show that the activation energy is lowest for K, which means it would react most vigorously. (note - these are not real activation energies - the atoms are not actually atomized to the gas phase, and the electrons are not removed under vacuum to infinite distance, but they still give us a good relative estimate)
However, knowing that the kinetics of the reaction depends on the activation energy won't help you if you don't happen to have a chart of activation energies handy. On an exam, you certainly won't, and it is good to develop an intuition for these sorts of things anyhow. And so, you might ask:
Why is the activation energy lower?
Looking at the links, you can see that the activation energies are dominated by the ionization energy. In other words, when it takes more energy to remove electrons, the reaction slows down.
Why should K have the lowest ionization energy? First we need to look at the equation that governs the potential energy between charges:
Here E is electrostatic potential energy, $\kappa$ is Coulomb's constant, and $Q_1$ and $Q_2$ are charges.
You can see from this that if the charges have the opposite sign, then energy will be become more negative as they move closer together (inversely proportional to distance). As either of the charges increase, the energy will become more negative as well (directly proportional to charge). In other words, big charges are harder to pull apart, and charges that are close together are harder to pull apart than charges that are far apart. As a rough analogy, imagine pulling apart two magnets - it is more difficult when they are close together.
Now compare K to the other group I element (Na). K is further down the column. Since atomic radius increases with the row number (due to the fact that orbital radius increases with principle quantum number n), the valence electron on K is further from the nucleus (on average) than it is on Na. This means that it is easier to remove a valence electron from K. As an analogy, imagine trying to move a rock up a hill. You have to do less work (put in less energy) if the rock is already halfway up the hill. Since the electrons are further from the nucleus in K, they are already "further up the hill."
What about group II elements? As you progress from left to right, the atomic radius decreases - this is because the effective nuclear charge seen by the valence electrons increases. The combination of increased charge and decreased distance leads to a steeper "hill" that must be overcome. On top of that, now we have to remove two electrons for the reaction to work, which is roughly twice as difficult (technically we don't have to remove both, but the overall reaction thermodynamics won't work out if we don't). As a result, group I elements will always react more vigorously with water than group II will.
I think that it helps to imagine reactions as sort of a "fight" between elements over electrons. Who will win the fight and how it will go is determined by the number of protons in the nucleus and the electron orbital configuration. This approach is useful because it applies to more advanced chemistry as well.
For some demonstrations of alkali metal reactions with water, check out this video: