[IKE]

Ok, in case this is stumping anyone this weekend, I will put my thoughts down in White:
So with the spring loaded kickers, most need a mechanical endstop to avoid penalty. Because of this, the ball will tend to follow a given trajectory. If you use trajectory physics, you can calculate the given distance if you know a couple parameters for this problem, ball speed and angle would give you distance. How? and How is velocity related to the distance? Well the lateral distance covered will be equal to the time in flight multiplied by the lateral component of the ball velocity. The time in flight is dictated by the vertical component of the ball velocity. Since gravity only slows the vertical component, 1/2*t=Vy/g with g being the gravity constant. Therefore the distance (d) traveled is d=Vx*t=Vx*2*Vy/g. This means that really distance is proportional to a constant (c) multiplied by the square of the velocity (since Vx and Vy have a constant relationship). Thus d is proportions to V^2.
So how do we find V? Well 1/2mV^2 is the kinetic energy imparted on the ball by the kicker. Where does this energy come from? The spring. Thus the spring energy 1/2K*x^2 is directly proportional to the ball kinetic energy 1/2*mV^2. From this we realise we want to maximize the kinetic energy.
Now for this particular problem, as many have noticed, the motor stalls thus it is torque limited thus Kx (spring force) is maximized. As we have seen though, energy is really what we are looking for. lets assume both K = 2 and X = 1 Then the energy stored is 1/2K*x^2=1/2*2*1^2=1 Now if we reduce the spring rate and hold the force constant Kx so that K = 1 and X = 2 then 1/2*1*2^2=4/2=2. We actually doubled the energy going into the kicker. This assumes though that you have the extra travel. what if you don't have the extra travel? Can weaker springs still give you more energy? Initially you would say no, if they both pull the same distance, then the stronger spring should store more energy. This assumes no preload though! so back to our earlier example. K=2 and X=1 gives us an energy of 1. Now with k =1, X can equal 2, but there isn't enough room. therefore we pre-stretch the spring so that it ends up with a total stretch of 2. Thus the energy for that is E= 1/2*K*(x2)^2-1/2*k*(x1)^2 where x2=2 and x1=1 thus E=1/2*k*(x2^2-x1^2) or E= 1/2*1*(2^2-1^2)=3/2 thus even in a restricted space, we still have 50% more energy with the weaker spring rate.

So my answer is B sometimes with springs, less rate is more.
Good luck to the teams competing this weekend.