Lead Compensator for double integrator
I got the position of the pole to be \$8.873\$ and the gain to be \$8.873\$.
First,

yields \$z = 0.8873\$. Next plug in \$z\$ into the transfer function above and you get \$-0.1127\$. The gain is then \$\frac{1}{0.1127} = 8.873\$.
Step response is as follows

I also had issues with the overshoot (about \$33 \%\$) but at least it meets the pole-zero relationship. I'm guessing it's an issue with the two poles at \$0\$.