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.
JDJ