To solve this problem we will apply the concepts related to the continuity equations. Here we will relate the input flow and the output flow depending on the cross-sectional area and the fluid velocity.
When we find the speed found, apply the kinematic equations of motion that relate the change in speed with acceleration (gravity) and distance traveled. So:
[tex]Q_1 = Q_2[/tex]
[tex]A_1 v_1 = A_2v_2[/tex]
Here,
[tex]A_{1,2}[/tex] = Cross sectional Area at each section
[tex]v_{1,2}[/tex] = Velocity at each section
For a circle we have that the relation would be,
[tex]\pi (r_1)^2 v_1 =\pi (r_2)^2 v_2[/tex]
[tex](r_1)^2 v_1 =(r_2)^2 v_2[/tex]
With the previous relation given as
[tex]r_2 = \frac{r_1}{2}[/tex]
We have replacing that,
[tex]v_1 = \frac{v_2}{4}[/tex]
[tex]v_2 = 4_v_1[/tex]
[tex]v_2 = 0.35*4[/tex]
[tex]v_2 = 1.4m/s[/tex]
From the cinematic equations of motion we have to
[tex]v_2^2-v_1^2 = 2ax[/tex]
[tex]1.4^2-0.35^2 = 2 (9.8)x[/tex]
[tex]x = 0.09375 m[/tex]
Therefore the distance will be 9.37cm
