1. The problem statement, all variables and given/known data A water bottle has a tap tube at the bottom. The diameter of the valve opening is 1 cm. If you turn on the tap, the water drains out with a velocity of 1.4 m / s. If the air pressure in the bottle is equal to the atmospheric pressure outside it, how high the water than in the bottle? (Assume that the water inside the bottle flows so slowly that you may neglect the speed.) 2. Relevant equations P+ρ×g×h+½×ρ×(v)2 = P+ρ×g×h+½×ρ×(v)2 3. The attempt at a solution Pressure outside is equal to inside so you can ignore it . So ρxgxh = ½×ρx(v)^2 1000 x 9.81 x h = ½x1000 x (1.4)^2 h= 9.990 cm I could solve this but what i don't understand is why is the dynamic pressure which is 1/2 x ρ x (v)^2 on the left side of the equation equal to zero ??? and why is the hydrostatic pressure which is ρxgxh equal to zero on the right side of the equation ?