Simple Fluid mechanics (bernoulli's?)

[mechEstudent]

Homework Statement

A can of Coca‐Cola has a small pinhole leak in it. The Coke sprays vertically
into the air at a height of 0.5 m. What is the pressure inside the can?
ρ = 1.11 g/mL

Homework Equations

Bernoulli: p1/ρ + (v1^2)/2 +g*z1 = p2/ρ + (v2^2)/2 +g*z2

The Attempt at a Solution

I've assumed:
z1=0, z2= 0.5
p2= atmospheric = 0 (gage)
v2=0

This simplifies bernoulli's to p1 = (g*z2-(v1^2)/2)/ρ
However v1 is still unknown and i have no way to solve for it

 

[rude man]

Homework Statement

A can of Coca‐Cola has a small pinhole leak in it. The Coke sprays vertically
into the air at a height of 0.5 m. What is the pressure inside the can?
ρ = 1.11 g/mL

Homework Equations

Bernoulli: p1/ρ + (v1^2)/2 +g*z1 = p2/ρ + (v2^2)/2 +g*z2

The Attempt at a Solution

I've assumed:
z1=0, z2= 0.5
p2= atmospheric = 0 (gage)
v2=0

This simplifies bernoulli's to p1 = (g*z2-(v1^2)/2)/ρ
However v1 is still unknown and i have no way to solve for it

Pretend you're a tiny bug just under the top surface of the can (where the hole is). How fast would you be moving? Or if you were at the bottom of the can?

 

[Camille]

Try to find $v_1$ from the simple equations of motion, in fact we have here the case of a vertical throw upwards.

 

[mechEstudent]

ah, i see now that v1=0 as well, since point 1 is to be taken inside the can. I was mistakenly assuming v1 to be an exit velocity. i knew it was something simple. Thank you

 

[rude man]

ah, i see now that v1=0 as well, since point 1 is to be taken inside the can. I was mistakenly assuming v1 to be an exit velocity. i knew it was something simple. Thank you

Correct! v1 = 0 everywhere inside the can. Inside the can, the pressure term in Bernoulli is exchanged for potential energy as we go from the bottom of the can to the top. But v1 = 0 everywhere inside.