a cylinder that is 20 cm tall is filled with water. If a hole is made in the side of the cylinder, 5cm below the top level, how far from the base of the cylinder will the stream land? Assume that the cylinder is large enough so that the level of the water in the cylinder does not drop significantly.
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
(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