Water Height of a centered-hole tank Here's the problem.

A large tank is filled with water for a height $H$. In the center of the tank a circular hole with a radius $R_1 = 10$ cm is made from which the water flows vertically. We can observe, at a distance $L = 50$ cm from the hole, the water jet has a circular section of radius $R_2 = 9$ cm. Assuming that the hole is small compared to the surface of the tank and therefore the speed, with which the surface of water is lowering, can be neglected, determine $H$.

I put here a quick-made image:

where red lines indicate data the problem gives me, yellow line what I've to find out.
Finding a solution:
At point $(1)$, the pressure applied is the atmospheric's, the same is applied at point $(3)$. So I can apply the continuity function:
$$A_1v_1=A_2v_2$$
I'm trying of using Bernoulli's equation:
$$P+\frac{1}{2}\rho v^2+\rho gh=constant$$
and if the pressures at $(1)$ and $(3)$ equal and because of $A_1<<A_3$, I get: 
$$v_1=\sqrt{2gh}$$
At this point, what should I do?
Additional question: I don't understand why the pressures at $(1)$ and $(3)$ equal and it's the atmospheric's. Is there not the pressure of water column in tank, applied in $(1)$?
 A: The reason for the equality of pressures in (1) and (3) is a simlifying assumption basing on that the water in the tube under the tank has far greater kinetic energy than in the tank (by continuity law a lot more velocity in the tube than in the tank). This kinetic energy from the outflow compensates the pressure term in Bernoulli's equation. Holds only at the hole, not on the bottom of the tank.
The calculation for $v_1$ was correct.
Now on (2) there is just the atmospheric pressure and thus, $P=0$ can be set. You will obtain another relation
$\frac{1}{2} \rho v_2 = \rho g (L+H)$ (Equilibrium between potential and kinetic energy).
Finally, use the equation of continuity and solve the equation for $H$.
A: With regard to the pressures at locations 1 and 2, if you look at the pressure at the surface of the free stream at these locations, it is atmospheric.  Because of Newton's 3rd law, this is the same pressure exerted in the liquid at these locations.  Now, since the radial acceleration of the fluid in the free stream is essentially zero over the cross section of the free stream, the pressure will not be varying with radial position.  Thus, over each of the cross sections, throughout the entire cross section, the pressure is atmospheric at locations 1 and 2.
