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If you have an open tank filled with water, and you poke a hole half way down the tank, you can solve for things like the speed of the water coming out of the hole with the Bernoulli's equation, $P_1 + \rho gh + \frac{1}{2}\rho v_1^2=P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2$
Why do these pressures cancel out?
If the tank is open, the surface of the liquid inside is exposed to atmospheric pressure. Same for the liquid flowing out the hole in the side.
