OK well despite you being undeserving of it let me explain some fundamentals of physics and flight to you. Lets say the aircraft touches down with a velocity, U. The distance required to bring the aircraft to a halt, s, is:
s=ut+0.5at^2
where a is the accelleration rate and t is the time.
The time is defined by the equation:
v=u+at,
where v is the final velocity (zero in this case), hence:
t=u/-a
The decelaration capabilty is of the aircraft is dependent on the capability of the brakes to slow the wheels and the tyres to grip the runway to exert a decelerating force. The decelerating force F, is determined by the coefficient of friction, mu, multiplied by the normal reaction of the aircraft N, which is the product of the mass of the aircraft m, times by the gravitational accelaration, g. Thus:
F = muN
OR
F = mumg
So working back we can say that:
S=u(u/-a)+0.5a(u/-a)^2
OR
S= -u^2/a +0.5u^2/a
Thus
S=-u^2/a
The decelaration force, F, comes primarily from the landing gear:
F=ma.
Thus
a=F/m
OR
a= mug
Thus
S= u^2/mug
So given that we are not landing on the moon, as you perceptively pointed out, g is constant. U is predetermined as the minimum speed required to keep the aircraft in the air. Thus the landing distance, S, varies according to the coefficient of friction, mu.
Wet runways have a lower coefficient of friction than dry runways (can you drive a car? Wet stopping distances are greater than dry stopping distances for that reason). When the runway is wet, mu is lower, S is higher and if S plus a safety margin exceeds the runway length then you cannot land on the runway. Now go away with a pen and paper, try working out some sums for yourself and leave the professionals to fly the aeroplanes.