How would I solve for the height of a ball on its second bounce?

The height a ball will reach on a bounce is directly proportional to the total energy in the ball (because at the top of the bounce KE=0 and PE = m*g*h, and only h can change).

If you lose 12% energy on each bounce, then the ball retains 88% energy each bounce as well.

E(0) = E

E(1) = E * (1 - .12)

E(1) = E * .88

E(2) = E * .88 * .88

E(n) = E * (.88)^n

So if release height is H,

H(0) = H

H(2) = H * .88^2

H(2) = .78 H

Suppose it reaches a height h after the 1st bounce

The total energy at that point is the gravitational potential energy = mgh

12% of the energy is then lost on the second bounce.

This leave 88% of mgh = 0.88mgh for the next highest point ( = mg(0.88h)

So the height after the second bounce = 0.88h

The height after the second bounce is 88% f the height of the first bounce.