# Water fountain problem

dronell90

## Homework Statement

Suppose the water fountain in fountain hills, Arizona, rises 150ft above the lake. Neglecting wind effects and minor losses, determine the velocity at which the water is ejected

## The Attempt at a Solution

SHISHKABOB
did you make an attempt at this problem?

dronell90
did you make an attempt at this problem?

yes, i tried using this formula, but i don't know if its right

A = 9.8m/s^2
V^2 = 0^2 + 2(9.8)D
V^2 = 19.6D

LawrenceC
V in your equation is final velocity which should be zero. Vo is initial velocity which is what you are seeking. The sign on A is negative 9.8. And don't mix feet and meters.

dronell90
I assume what you mean is that water is squirted directly upward so it reaches a peak 150 ft above the jet where it exited. If that is the situation, it is the same as shooting a projectile straight up.

yes that's what i meant. is the steps i used in the above right?

SHISHKABOB
the problem is asking for the *initial velocity*, which is v0 in that formula

NasuSama
yes, i tried using this formula, but i don't know if its right

A = 9.8m/s^2
V^2 = 0^2 + 2(9.8)D
V^2 = 19.6D

Nope, the initial velocity is not right. The water rises up to 150 ft. What does this tell you?

..and also, the given unit is ft. Then, g = 32 ft/s/s

dronell90
yes, i tried using this formula, but i don't know if its right

A = 9.8m/s^2
V^2 = 0^2 + 2(9.8)D
V^2 = 19.6D

Nope, the initial velocity is not right. The water rises up to 150 ft. What does this tell you?

..and also, the given unit is ft. Then, g = 32 ft/s/s

So it should be like this
A = 32 ft/s
V^2 = Vo^2 + 2(32)150

then what is the value fo V^2

SHISHKABOB
when you throw something up into the air, there's a point at which it reaches its max height and then starts to come back down, yes? What's the velocity at the *instant* it reaches its maximum height?

NasuSama
So it should be like this