Did I do this fluids problem right? A find the gallons/min gushed by a fountain

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The discussion focuses on calculating the flow rate of water from a fountain that reaches a height of 4.23 m, using the effective area of the pipe. The original attempt involved incorrect application of formulas, particularly the centripetal force equation, which is not relevant to this problem. A suggestion was made to use energy conservation principles instead, which could provide a more accurate approach to solving the problem. The user expressed frustration after multiple unsuccessful attempts. The conversation emphasizes the importance of using appropriate physics concepts for fluid dynamics calculations.
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1. A fountain sends a stream of water straight up into the air to a maximum height of 4.23 m. The effective area of the pipe feeding the fountain is 5.38 x 10^-4 m^2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10^-3 m^3)



Homework Equations



F = (mv^2)/r
F = P * A
P = density * g * h
m = density*(Ah)

The Attempt at a Solution



P= 1.013x10^5 * pgh
P= 1000*9.8*4.23 + 1.013x10^5
P= 142754 Pa

PA = F
PA = (mv^2)/r

r = (Area/pi)^(1/2)
r= .013086 m

mass = 1000 * area * 4.23
mass = 2.27574 kg

142754 Pa * (5.38 *10^-4 m^2) = (2.27574 kg * v^2)/.013086 m
v = .66455 (m/s multiply by area

.66455 * (5.38 * 10^-4) = .0001926 m^3/s

divide v by 3.79 * 10^-3 m^3, then multiply by 60 seconds

5.66 gal/min



this is so obviously wrong, but I've tried this prob beyond five times already and am at wits end... can anyone steer me in the right direction? thankkkks!
 
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lalalah said:
F = (mv^2)/r

That equation isn't relevant here, since it applies to circular motion and centripetal force.

How about trying energy conservation?
 
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