Answer
The second vaulter clears the bar at $1.67m/s$.
Work Step by Step
Take the point where the vaulter clears the bar to be the starting point and where the vaulter lands to be the final point.
During the fall, the vaulter's KE increases while PE decreases by the same amount, following the principle of energy conservation. Therefore, $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2} (v_f^2-v_0^2)=-g\Delta h$$
For both vaulters, the bar's height $\Delta h$ is the same. The differences are in $v_0$ and $v_f$. Therefore, $$\frac{1}{2} (v_{1, f}^2-v_{1, 0}^2)=\frac{1}{2} (v_{2, f}^2-v_{2, 0}^2)$$ $$v_{1, f}^2-v_{1, 0}^2=v_{2, f}^2-v_{2, 0}^2$$ $$v_{2, 0}=\sqrt{v_{2, f}^2+v_{1, 0}^2-v_{1, f}^2}$$
The first vaulter clears the bar at $1m/s$ and lands at $8.9m/s$, while the second vaulter lands at $9m/s$. So $$v_{2, 0}=\sqrt{9^2+1^2-8.9^2}=1.67$$
