Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
which is 1 cm wide.
How wide is the stream 30 cm below the tap (assuming that the stream
does not break up due to surface tension induced instability).
No one bit on this problem, so I will give my farthings worth.
I think I have seen a problem like this before, a while back.
I think the width would be proportional to how far it is from the tap.
If I remember right the formula is
Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.
We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume
Because of gravity and all that, the rate this disk is passing y is:
Where F is the rate the water is leaving the tap.
Now, solve for R(y).
Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.
This does look like an interesting little problem, though.
The mass flow rate is a constant so at any point Av is that same
as at any other point, where A is the crossectional area of the stream
and v is the speed of the stream.
A mass element undergoes an accelaration of g m/s^2 (sign convention
is positive downwards), and so at time t its speed is:
and it has fallen a distance:
So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):
the roots of this are t~=-0.2685, and t~=0.22776. The first
of these roots is unphysical so we are left with t~=0.22776 seconds.
Its speed when it has fallen 0.3 m is:
v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s
So now we have from the constancy of the mass flow rate past any point:
D~=0.002872 m, or D~=0.29 cm.