1. The problem statement, all variables and given/known data Two water jets collide and form one homogeneous jet as described below: Jet 1, with v=4 m/s, enters directly upwards through a pipe of diameter .10m into a connector pipe. Jet 2, with v= 6 m/s, enters from the left in a pipe of .12m diameter which is perpendicular to Jet 1. Both jets meet and continue down one pipe. Find the discharge of homogeneous fluid at the end of the pipe, and find the horizontal and vertical components of the fluid in the third pipe. 2. Relevant equations The continuity euation...Q=A1V1=A2V2. And possibly the conservation of momentum equation: F= pQ2V2^2 - pQ1V1^2 3. The attempt at a solution The first part is quite easy...A=pi*(d^2)/4...v*A=V (volumetric flow rate) add the two streams together to get the discharge out the third pipe. So A1*V1+A2*V2= pi*(.12^2)/4 * 6 m/s + pi*(.1^2)/4 * 4 m/s = 0.100 m^3/s Now, I have no idea where to begin. I don't have an angle for the third pipe which (in the diagram) veers at an angle theta from the joined 1st and 2nd pipes. I also have the answers as this is a practice exam for my exam tomorrow. Horizontal Component: 4.1 m/s Vertical Component: 1.3 m/s Theta: 17.6 degrees (not given) Thanks for all the help.