Answer:
40.7 m
Explanation:
Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:
[tex]z_1=16\ m,P_1=2\ atm,V_1=0(velocity\ at\ the \ surface \ of\ water\ is\ low)\\V_2=0,P_2=P\\\\Using\ Bernoulli\ equation:\\\\\frac{P_1}{\rho g} +\frac{V_1^2}{2g}+z_1= \frac{P_2}{\rho g} +\frac{V_2^2}{2g}+z_2\\\\Sine\ V_1=0,V_2=0:\\\\\frac{P_1}{\rho g} +z_1=\frac{P}{\rho g} +z_2\\\\z_2=\frac{P_1}{\rho g} -\frac{P}{\rho g} +z_1\\\\z_2=\frac{P_1-P}{\rho g} +z_1\\\\z_2=\frac{P_{1.gage}}{\rho g} +z_1\\\\[/tex]
[tex]z_2=\frac{2\ atm}{1000\ kg/m^3*9.81\ m/s^2}(\frac{101325\ N/m^2}{1\ atm} )(\frac{1\ kg.m/s^2}{1 \ N} ) +20\\\\z_2=20.7+20\\\\z_2=40.7\ m[/tex]
