Tap

As we are treating the water as an incompressible fluid, it is true that for any two parts of its trajectory $A_1 v_1 = A_2 v_2,$ where $A_i$ is the cross-section area at the hight where its velocity is $v_i.$ If we take $A_1 = \pi r^{2}(0),$ then $v_1 = v_0.$ So we may write

$A_2 = \pi r^{2} = A_1 \frac{v_1}{v_2} = \pi r^{2}(0) \frac{v_0}{v_2}.$

But according to Torricelli, if we drop a distance $h$ then $v_2^{2} = v_0^{2} + 2gh.$ Therefore we may write

$r^{2}(h) = r^{2}(0) \frac{v_0}{\sqrt{v_0^{2} + 2gh}}.$

Simplifying

$r(h) = \frac{r(0)}{ \left(1 + \frac{2gh}{v_0^{2}} \right)^{1/4} }.$