# Water fountain problem

## 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

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did you make an attempt at this problem?

did you make an attempt at this problem?
yes, i tried using this formula, but i dont know if its right

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

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.

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 thats what i meant. is the steps i used in the above right?

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

yes, i tried using this formula, but i dont 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

yes, i tried using this formula, but i dont 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

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?

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
Well, gravity drags the "projectile of the water" down. Consider throwing up the ball for example. If you toss the ball, then you exert the initial velocity v_0. At the highest point (or at the max height), the ball stops rising. That means v = 0 ft/s or v_f = 0 ft/s. After that time, the ball falls down due to gravity.

Following the example I have shown to you, find v_0.