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

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The forum discussion centers on calculating the flow rate of water from a fountain that reaches a maximum height of 4.23 meters, using the effective area of the pipe at 5.38 x 10^-4 m². The user attempted to apply various equations, including pressure and force equations, but struggled to arrive at the correct gallons per minute output. The correct approach involves using energy conservation principles rather than centripetal force equations, leading to a calculated flow rate of approximately 5.66 gallons per minute.

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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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