1. The problem statement, all variables and given/known data I just need someone to check my work. (If what I'm doing is right, if not, tell me where I have gone wrong, and what I should do) Please see attachment for the question. 2. Relevant equations P + 0.5[tex]\rho[/tex]v^2 + [tex]\rho[/tex]gh = constant [tex]\rho[/tex]water = 1kg/m^3 3. The attempt at a solution I'm not too sure if the pressure is different from inside the water tank to the outside. The question is asking for the speed at which water leaves the spout, for is that the final speed? I have assumed the pressure is the same inside and out, please tell me if I'm wrong. The density varies with temperature, and the temperature inside the water tank and the temperature outside the water tank won't be the same, but I have assumed them to be the same, so the density will be the same on either side of the equation below. P + 0.5[tex]\rho[/tex]v^2 + [tex]\rho[/tex]gh = P + 0.5[tex]\rho[/tex]v'^2 + [tex]\rho[/tex]gh' Pressure cancels out, but the rest do not as v and h are not the same on each side. 0.5[tex]\rho[/tex]v^2 + [tex]\rho[/tex]gh = 0.5[tex]\rho[/tex]v'^2 + [tex]\rho[/tex]gh' 0.5v^2 + gh = 0.5v'^2 + gh' I think the initial speed is 0, because I wouldn't know what other value it might be, again, please tell me if I'm wrong! gh = 0.5v'^2 + gh' 9.8 x 12 = 0.5v'^2 + 9.8 x 4 78.4 = 0.5v'^2 156.8 = v'^2 So v' = 12.52 m/s For the second part of the question, b) isn't it just h=0? For the water to hit the ground as far as possible from the base of the tank, the height at which the spout should be placed is 0m, isn't it?