According to your question, we are finding the height of second bounce.

If I say it's** s,**** **

the u(initial velocity) of first bounce should be 0, because you said you are dropping, not adding force.

the v(final velocity) of first bounce should be u(initial velocity) of second bounce.

and the ball at peak of second bounce, the speed should be 0.

if we work out the acceleration by F=ma, (or if time or acceleration is given and you didn't mention) then we can work out s. By formula $${{\mathtt{v}}}^{{\mathtt{2}}} = {{\mathtt{u}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{as}}$$

$$0=(u_1) ^2+2as$$ (I'm setting peak of second bounce as end)