Could anyone offer me some guidance on parts b and c of the following question? The pressure in a uniform pipe of flowing water is enough to hold up 4.00 m of water in an open, vertical tube as shown in Figure P.60. Assume that water is an ideal fluid. The pipe leads 3.00 m upstairs and empties into the open air through an opening with an area one-fourth that of the uniform pipe. (a) At the exit opening of the pipe, what is the pressure? (b) What is the pressure at the point where the vertical tube is connected to the pipe? (c) What is the exit speed of the water from the pipe? this is the picture of the problem: http://img346.imageshack.us/img346/5326/p11607ht.gif" for part (a), i figured b/c the water is open to the air, its pressure is 1 atm (approx. 1.01E5 Pa). for parts (b) and (c), i think i need to somehow use the equation: P1 + (1/2)*density*v1^2 + density*g*y1 = P2 + (1/2)*density*v2^2 + density*g*y2 but i'm not sure how to get the velocity or pressure at the point under the 4m opening. any ideas/hints for me? thanks!!