## College Physics (4th Edition)

The top of the water in the tower is $8.6~meters$ higher than the faucet.
We can find the speed $v_2$ of the water when it comes out of the faucet: $A~v = \frac{V}{t}$ $v = \frac{V}{A~t}$ $v = \frac{(\pi)(0.22~m)^2(0.52~m)}{(\pi)(0.0127~m)^2~(12~s)}$ $v = 13.0~m/s$ Note that the pressure at the top of the water in the tower and the pressure outside of the faucet are both equal to the atmospheric pressure. We can use Bernoulli's equation to find the height $h_1$ at the top of the water in the tower: $P_1 + \rho~g~h_1+\frac{1}{2}\rho~v_1^2 = P_2+\rho~g~h_2+\frac{1}{2}\rho~v_2^2$ $\rho~g~h_1+0 = 0+\frac{1}{2}\rho~v_2^2$ $h_1 = \frac{v_2^2}{2g}$ $h_1 = \frac{(13.0~m/s)^2}{(2)(9.80~m/s^2)}$ $h_1 = 8.6~m$ The top of the water in the tower is $8.6~meters$ higher than the faucet.