Find the shape of flowing water using Torricelli's law

Question

Suppose I have a cylinder tank with R radius and height h, and water filled with water up to the level h. There is a vertical rectangular slot on the surface with sizes h x 1 cm. There is only gravitation, no air resistance etc.

Obviously, a jet flowing from the small hole will have parabolic trajectory. But I want to define, what shape has the water flowing from the slot. I assume that vertical slope is like aligned holes with no distance betwwn them. Will the shape be like rectangular triangle with 45 degrees angles (like on 1st picture below) or it will have parabolic shape (like on 2nd picture below)?

Answer 1

I am going to assume for simplicity that the exit velocity is given by Toricelli’s law: $$ v=\sqrt{2gh} $$ Taking cartesian coordinates $x,y$ horizatoal and vertical respectively with the origin at the top to the container. I therefore have: $$ h=-y $$ Assuming freefall, the trajectory starting at $y=-h$ has equation: $$ x= \sqrt{2gh}t \\ y=-h-\frac{gt^2}{2} \\ y=-h-\frac{x^2}{4h} $$ You therefore have a family of curves parametrized by $h$. You now just need to do the math to find the envelope.

Your first intuition is correct. You can see it by noticing that (completing the square or the AGI): $$ y\leq x $$ with equality at $x=2h$. Thus the envelope is the 45° line.

Hope this helps.