Superballs have a coefficient of restitution of .92.
COR= sqrt(bounce height/drop height) against a perfectly hard surface.
So, it should bounce to about 85% of its drop height. A 1 meter drop, then, would result in a bounce of 85cm.
Of course, we need to account for air.
The terminal velocity of a superball is pretty easy to calculate:
Vt= sqrt(2mass g\DensityOfAirAreaCoefficientOfDrag)
The drag coefficient for a smooth sphere is about .5, the density of air is 1.29kg/m^3. I don’t know the density of a superball, but it’s probably a little more dense than water, so let’s guesstimate a 3.0 cm ball with waterish density.
The terminal velocity would then be about 34 m/s, or 75 mph.
That’s about equal to a fall of 250m. (I’m not showing my math on this one, because I used my calculator’s formula.) So the highest bounce from a dropped ball would be about 215m, give or take, depending the assumptions.
BUT - first we need to know how much force it takes to start permanently deforming the ball (cracking, bursting, etc). I’d wager it is more than the force you’d get at 75 mph.
If the ball breaks below 75 mph, then your max bounce would be lower. If it breaks at more than 75, the max bounce is obviously going to be higher, but you need some sort of way to get the ball moving faster than terminal velocity. The easiest and pretty much just as accurate way is just to ignore the effect of the atmosphere, figure out the maximum impact the ball can withstand, then figure out the height of your bounce from there.
Interestingly, glass spheres have a COR of .95 – way more than that of a superball. Of course, glass spheres tend to break more easily than a rubber ball, so the effect is harder to see at normal scales.