Now, considering that the area of position one, \(A_1\) is so large that \(v_1\) decreases and approaches zero. Additionally, \(h_2\) is located at position 2, which is our base zero point, so \(h_2\) equals zero. Both \(P_1\) and \(P_2\) are 1 atm because they are both exposed to air, and so \(\Delta P\) amounts to zero also. Thus, we are left with:

\[0 = \rho \textit gh + \frac {\rho (\textit {-v}_2^2)}{2}\]

\[\frac {\rho (\textit {v}_2^2)}{2} = \rho \textit gh\]

\[\frac {\textit {v}_2^2}{2} = \textit gh\]

\[\textit {v}_2^2 = \textit 2gh\]

Thus effectively yielding Torricelli’s Equation:

\[\textit {v}_2 = \sqrt{\textit 2gh}\]