Velocity of Fountain Water


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.


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 dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know.