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