- #1

dronell90

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## 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

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- Thread starter dronell90
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- #1

dronell90

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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

SHISHKABOB

- 541

- 1

did you make an attempt at this problem?

- #3

dronell90

- 6

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did you make an attempt at this problem?

yes, i tried using this formula, but i don't 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

- #4

LawrenceC

- 1,198

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- #5

dronell90

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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?

- #6

SHISHKABOB

- 541

- 1

the problem is asking for the *initial velocity*, which is v_{0} in that formula

- #7

NasuSama

- 326

- 3

yes, i tried using this formula, but i don't 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

- #8

dronell90

- 6

- 0

yes, i tried using this formula, but i don't 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

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

- #9

SHISHKABOB

- 541

- 1

- #10

NasuSama

- 326

- 3

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.

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