As others have highlighted, the question is with what ground speed you touch down. You might think that plus/minus a few knots doesn't really matter, but:
- your kinetic energy is quadratic in speed: E = 1/2 m v^2.
- if braking acceleration is constant, landing distance is quadratic in speed: s = v^2 / 2a
So, to adapt voretaq7's example, if your approach air speed is 45 knots, with a 15 knots headwind you land with 30 knots, while with a tailwind you land with 60 knots - now you touch down with four times the energy you need to dissipate, and need four times as much runway to stop (which, happily, lines up nicely with his POH numbers).
Now, finally, one more example:
Suppose the runway is such that with no wind, you can land with approach speed v and just stop prior to the end.
With a (mild) headwind of 10% of your airspeed, you touch down with 0.9v, thus you now use only 81% of the available runway, and have gained quite a buffer.
Whereas, if you have a 10% tailwind, you touch down with 1.1v. Thus, your required stopping distance now is 1.21 times the available runway. When you hit the end now, you'd need 21% of the runway length extra to stop. Now, this gets a bit tricky, but feel free to do the maths, this means that you hit the wall at the end with nearly half of your approach speed, still having about 21% of your (1v) landing energy left to dissipate.
So - what would you rather have, a 20% buffer of runway remaining (with head wind), or hitting the end of the runway with over 40% of your landing speed (with tail wind)? :-)