Answer
$143.5\ rpm$.
Work Step by Step
1. Based on the travel distance, we have $\frac{v_0^2}{g}=83.19$, thus $v_0=\sqrt {83.19\times9.8}\ m/s$.
2. Assume the rpm is $\omega$ ($\frac{\omega}{60}$ rotation per second), with a rotation radius $r=190cm=1.9m$, the linear speed is given by $2\pi r(\frac{\omega}{60})=v_0$, thus $\omega=\frac{60v_0}{2\pi r}=\frac{60\sqrt {83.19\times9.8}}{2\pi (1.9)}\approx143.5\ rpm$.
