1. The problem statement, all variables and given/known data " A water filled verticle pipe of cross section area 25 cm^2 is open at the top. If an outlet of cross section area 15 cm^2 is made on it's side at a depth of 10 m from the level of the water at the top, then the velocity of the water exiting from the outlet is? 2. Relevant equations " P1 + ρgh1 + 0.5 ρv1^2 = P2 + ρgh2 + 0.5 ρv2^2 . 3. The attempt at a solution Ok so what made me more confused is that v1 doesn't equal 0 because I was about to use this equation : √2gh Then I tried to find both of v1 and v2 with the equation of A1 v1 = A2 v2 and I got this " v2 = 1.667 v1" which didn't help much because it is already obvious that the velocity will increase since the area is smaller. After that I started cancelling some of the equation ( the one in brackets are cancelled ) : (Patm)+ ρgh1 + 0.5 ρv1^2 = (Patm) + (ρgh2) + 0.5 ρv2^2 And I tried to plug in numbers : 1000 • 9.8 • 10 + 0.5 • 1000 v1^2 = 0.5 • 1000 v2^2 98000 + 500 v1^2 = 500 v2^2 And I was left blank. So I'm now thinking whether the cross section area has anything to do with the pressure... But yeah that was my tries .