Answer
The center of mass of the drops will be above the halfway point.
Work Step by Step
Take 2 parts of dripping water: part A above the halfway point and part B below the halfway point.
We have $$v_{cm}=\frac{m_Av_A+m_Bv_B}{m_A+m_B}$$
Take $m_A+m_B$ to be constant, while $v_{cm}$ is already always constant.
Now we know that the water's speed increases as it falls downward, so we can expect that $v_B\gt v_A$. Therefore, it follows that $m_B\lt m_A$, or there are more masses of water clustering in the upper part of the flow.
So the center of mass of the drops will be above the halfway point.
