Solving Fluid Problems: Distance of Water Drop

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SUMMARY

The discussion focuses on calculating the distance below a faucet where the radius of a water stream reduces to half its initial value. The initial speed of the water is 0.500 m/s. The correct approach involves applying Bernoulli's equation and the principle of conservation of mass, specifically the equation A1v1 = A2v2 = Q. By substituting the known values and solving for height (h), the problem can be resolved accurately.

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Fluid Problems :(

1. Water is running out of a faucet, falling straight down, with an initial speed of 0.500 m/s. At what
distance below the faucet is the radius of the stream reduced to one-half its value at the faucet?

I tried doing this by using Bernoullis equation and replacing r with 1/2r for the bottom part of the drop. Initially i canceled the pressures (since they are same all points) and assumed the initial height to be zero therefore giving 1/2v^2= 1/2vdrop^2+gh and got stuck after that


Any help is much appreciated
 
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Hi dreamx20, welcome to PF.
1/2v^2= 1/2vdrop^2+gh
The above equation is correct.
At every point of flow, the rate liquid flow is constant.
So A1v1 = A2v2 = Q
Substitute the values of v1 and v2 and solve for h.
 

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