Homework Help: Water fountain problem

dronell90 · Sep 23, 2012

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. The attempt at a solution

SHISHKABOB · Sep 23, 2012

did you make an attempt at this problem?

dronell90 · Sep 23, 2012

SHISHKABOB said: ↑

did you make an attempt at this problem?

yes, i tried using this formula, but i dont know if its right

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

LawrenceC · Sep 23, 2012

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 · Sep 23, 2012

LawrenceC said: ↑

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?

SHISHKABOB · Sep 23, 2012

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

NasuSama · Sep 23, 2012

dronell90 said: ↑

yes, i tried using this formula, but i dont know if its right

V^2 = Vo^2 + 2AD
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 · Sep 23, 2012

dronell90 said: ↑

yes, i tried using this formula, but i dont know if its right

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

NasuSama said: ↑

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
V^2 = Vo^2 + 2AD
A = 32 ft/s
V^2 = Vo^2 + 2(32)150

then what is the value fo V^2

SHISHKABOB · Sep 23, 2012

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 · Sep 23, 2012

dronell90 said: ↑

So it should be like this
V^2 = Vo^2 + 2AD
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.