I have a physics question for you...

I'm claiming "not enough information." We not only need to know how much energy is stored in the spring, we need to know how fast it is released.

If I knew that, I would still be waiting for someone else to answer.

First you have to remember that energy is always conserved an second you have to remember that fpf is a confusinf system as units of force (pounds) and mass (pounds) aren't the same--they're dictated by F=ma

Now lets get to basics

Springs are governed in genefal by F=kx where k is a spring constant and x is displacement, and the potential energy (PE) in a spring is given by.

PE=1/2kx^2

When I rearrange to get the PE from what you gave (k=F/x), you find

PE=1/2Fx putting in values an converting to a fps energy term (ft lbf)

PE=1/2*(100/16 lbf)*(5/12 ft)~1.30 ft-lbf

Now when you put a 1 oz projectile on this loaded gun you're limited to this amount of energy.

Not lets convert this PE to KE

KE=1/2mv^2

now m=1/16 lbm (lbm stands for pounds mass)

1.30 ft-lbf=1/2 (1/16 lbm)*v^2

but an lbm is mot the same as a lbf so a simple conversion isn't possible

Remember that F=ma and under an a of 1g (32 ft/s^2) 1lbm =1lbf

and 1 lbf= 32 lbm ft/s^2

so
1.30 ft 32 lbm ft/s^2=1/2*(1/16) lbm *v^2

cancelling everything then

1.30 (32^2) ft^2/s^2=v^2

taking thre square root to find the projectile velocity we find

v=32*sqrt (1.3) ft/s ~36.5 ft/s

Now we know that the 1 oz projectile is throun upward under 1 g of acceleration at 36.5 ft/s

now lets use Newtons laws of motion to find the maximum height of the projectile

at some time the projectile has zero upward velocity--what's this time

this is easy

v1=v0-at set v1=0, v0=36.5 ft/s , and a=32 ft/s^2
so
t=36.5 ft/s/(32 ft/s^2)~1.14 s

now we know as 1.14 s the projectile reaches its maximum elevation

what's this elevation

s=1/2at^2=1/2*(32 ft/s^2)*(1.14 s)^2~20.8 ft ~ 250 inches

so with a mass ratio of 100, your displacement ratio is 50.

Rap