How can I find water velocity between two cross-sections using continuity and Bernoulli?

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Taylan
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Homework Statement
Water is coming out of a tap. The surface area of water at A0=1,2cm^2 and at A=0,35cm^2. There is a distance of h=45mm between the two points ( see the attachment). g=9,81m/s^2.

a) What is the volumetric flow rate (Q) of water?
b) if the Q=0,3l/s, what would be the surface area of A? ( given that A0 and h are the same)
Relevant Equations
Q=V/t
Q= Av
A1v1=A2v2
a) so Q=V/t = Av
Q is constant ( same At A0 and A). That means I have to find the velocity of the water either at A0 or at A. But how can I find it? I thought there must be a reason that h is given but the only way I can use it is to assume the velocity is 0m/s at A0 but actually it is not. Bernouli equation wouldn't work for the same reason. Any tips?

Finding the distance for the water to flow from point A0 to A would also help but I see no solution for that time as well.
 

Attachments

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Discussion

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The water is in free fall. If ##v_0## is the velocity at the tap, from the appropriate SUVAT equation, what is the velocity (in terms of ##v_0##) at a distance h below the tap?
 
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Chestermiller said:
The water is in free fall. If ##v_0## is the velocity at the tap, from the appropriate SUVAT equation, what is the velocity (in terms of ##v_0##) at a distance h below the tap?

The thing is h gives the distance between A0 and A. However A0 is not where the water starts flowing. So at A0, it already has some velocity. ( that is what it looks to me from the attachment)
 
Taylan said:
The thing is h gives the distance between A0 and A. However A0 is not where the water starts flowing. So at A0, it already has some velocity. ( that is what it looks to me from the attachment)
Read my response again. Which SUVAT equation do you think is applicable?
 
Chestermiller said:
Read my response again. Which SUVAT equation do you think is applicable?

v^2 = u^2 + 2as
 
Taylan said:
v^2 = u^2 + 2as
Excellent. Now let u be the velocity at the tap (as yet unknown) and let v be the velocity h below the tap. OK so far?
 
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Chestermiller said:
Excellent. Now let u be the velocity at the tap (as yet unknown) and let v be the velocity h below the tap. OK so far?
yes
 
Taylan said:
yes
OK. Now u and v also have to satisfy another equation. That is the constancy of volumetric flow rate equation. What is that equation?
 
Chestermiller said:
OK. Now u and v also have to satisfy another equation. That is the constancy of volumetric flow rate equation. What is that equation?

A0.v0 = A.v .. and then setting up simultaneous equations?
 
Yes. Incidentally, the "free-fall equation" also follows from the Bernoulli equation:

$$\rho \frac{u^2}{2}+\rho g h=\rho \frac{v^2}{2}+0$$
 
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Chestermiller said:
Yes. Incidentally, the "free-fall equation" also follows from the Bernoulli equation:

$$\rho \frac{u^2}{2}+\rho g h=\rho \frac{v^2}{2}+0$$

Thanks a lot for the help!