Answer
The proof is below.
Work Step by Step
From the Bernoulli equation, we know:
$ P_B + 0 $ and $P_A+\frac{1}{2}\rho v^2$ are both equal. Thus, we find:
$P_B =P_A+\frac{1}{2}\rho v^2 \\ P_B -P_A=\frac{1}{2}\rho v^2 \\\Delta P =\frac{1}{2}\rho v^2 \\ v = \sqrt{\frac{2\Delta P}{\rho}}$
