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October 23, 2014

October 23, 2014

Posted by **zoe** on Sunday, December 2, 2012 at 2:38pm.

what do we know to plug in for bernouilli's equation?

- physics (SOS!) -
**Damon**, Sunday, December 2, 2012 at 3:31pm(1/2) rho v^2 + rho g z + p = constant

air pressure p is the same at the surface as when the water exits .05 meters below the surface so comparing conditions at the surface and at the exit:

0 + rho g (.2) + p air = (1/2) rho v^2 + rho g (.15) + p air

or

(1/2) rho v^2 = rho g (.05)

v = sqrt (2 g *.05)

In other words the water speed will be the same as that of an object that is dropped 5 centimeters.

v is our horizontal speed

v = sqrt (2 * 9.81*.05)

= 22 m/s

= 1 m/s

now how long to fall .15 meters?

.15 = (1/2) g t^2

.15 = (.5)(9.81) t^2

t = .175 s

so 1 * .175 = .175 meters from the cylinder

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