Two holes with an area $0.2cm^2$, each are drilled one above the other in the wall of a vertical vessel filled with water. The distance between the holes is $H=50cm$. Every second $Q=140cm^3$ of water is poured into the vessel. The stream flowing out of the holes intersect at a point. $x$ is the horizontal distance of this point from the vessel and $y$ is the vertical distance from the water surface in the tank. $x$ and $y$ remain same as the time passes. Find $x$ and $y$
This is a physics problem, but it involves quite a bit of math and kinematics. I think it’s okay to be asked on Math, since Physics SE doesn’t accept such questions.
I am making an assumption that area of holes $A_1, A_2<<< A$, where $A$ is area of cross section of tank.
From the continuity equation $$Q=A_1v_1+A_2v_2$$ $$140 =0.2(v_1+v_2)$$ $$v_1+v_2=700$$
Also, $v_1=\sqrt {2gh_1}$ and $v_2=\sqrt {2gh_2}$
$$v_2-v_1=\sqrt {2g} (\sqrt {h_2}-\sqrt {h_1})$$
We know that $h_2-h_1=50cm$
But that isn’t helpful in this case. If we just find $v_1$ and $v_2$, finding x and y should be simple eng(hopefully), but this is where I am stuck at.
Note that the speeds with which the water leaves the holes are proportional to the water pressures at the holes, which are proportional to the vertical distances between the holes and the water surface. Let $h$ be the distance between the upper hole and the water surface. Then
$$\frac{v_1}{v_2} = \frac h{h+50}\tag 1$$
Also, the kinetic energy with which the water leaves the holes is converted from the difference of gravitational potential energy between the holes and the water surface, that is,
$$\frac12 \Delta m_1v_1^2 + \frac12 \Delta m_2v_2^2 = \Delta m_1gh + \Delta m_2g(h+50)$$
where $\Delta m_1$ and $\Delta m_2$ are the amount of water leaving the two holes, respectively, which are proportional to their speeds leaving the holes. So, rewrite the energy equation above as,
$$\frac12 v_1^3 + \frac12 v_2^3 = v_1gh + v_2g(h+50)\tag 2$$
Then, along with the equation you already had,
$$v_1+v_2=700\tag3$$
the equations (1), (2) and (3) have three unknowns in $v_1$, $v_2$ and $h$ and they can be solved from the system of the three equation derived.
As you indicated, once you obtain $v_1$, $v_2$ and $h$, finding the cross point $x$ and $y$ should be straightforward by examining the intersection of the two projectile trajectories.