Velocity of Fountain Water

## Question:

I am a freshman at my community college. i am currently in a beginning type
physics class. we were asked by my professor to find a way that we can
measure the velocity of water exiting a drinking fountain. and i have NO
CLUE!!! please help me. what he means is when you press the button to turn
the drinking fountain on, how fast does the water leave the little hole. and
then he also probly wants to know how long it takes to hit the actual
fountain. if you could help me out i would greatly appreciate it. thank you
so much.
## Answer:

The kinetic energy of the water leaving the nozzle is converted to potential
energy as it rises. By measuring the height to which the water rises after
leaving the nozzle you may find the potential energy at the top of its path
when its velocity is about zero. Potential energy is given by the formula
mgh when m is the mass of the object, g is the acceleration due to gravity
and is the height above the reference point.
Let's assume that the energy of the water is all kinetic at the nozzle exit.
Kinetic energy is given by the formula 1/2mv^2. Since energy is conserved,
the kinetic energy at the nozzle will all be converted to potential at the
top of the stream. Therefore we can set the potential energy at the top
equal to the kinetic energy at the nozzle. mgh=1/2mv^2. The m's cancel out
so that gh=1/2v^2 or v^2=2gh or v=square root of 2gh. The value of g is 9.8
m/s/s and h should be measured in meters.

To find how long it takes the water to fall from the top of its path to the
fountain we can just use s=1/2gt^2, where s is the distance fron the top of
the stream to the fountain basin and solve for t. Likewise on its way up
the water is subject to gravitational acceleration so it will reach the top
of the stream when h=the velocity you found in part 1, call it v, times
t-1/2gt^2 or h=vt-1/2gt^2. Again solve for t.

This information is brought to you by M. Casco Associates, a company
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JDJ