Water fountain height 4.23 m, find flow rate in gallons per minute

[lalalah]

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!

 

[Redbelly98]

F = (mv^2)/r

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

How about trying energy conservation?