How Does Water Jet Force Affect Reservoir Pressure Calculations?

[gaugie]

Homework Statement

A reservoir filled with water to a height h has a hole with area A on height h1 with h1 < h.
What is the force on the reservoir? I neglected the fact, that the height of the reservoir itself changes.

The Attempt at a Solution

I started with Bernoulli at the top and at the hole with p_a being the atmospheric pressure:
$$p_a + \rho g h =c$$ ...
$$p_a + \rho g h_1 +\frac{1}{2} \rho v_1 ^2 =c$$.
With this I get:
$$v_1 ^2 = 2g(h-h_1)$$
Now I want the Force, with I being the impuls:
$$F=\frac{dI}{dt}=\frac{dV}{dt}\rho v_1=Av_1\rho v_1=A\rho v_1 ^2$$
and therefore I get:
$$F=2gA\rho(h-h_1)$$

I now would want to know, if this solution is correct and furthermore, why I can't just assume that the force is due to the lack of pressure on one side and therefore just the pressure on height h1 times area A on the opposite side, but rather get two times this force with my solution.
Thanks in advance,
gaugie

edit: the post should actually be in the "introductory physics" forum, but was accidentally postet here. I don't know, if it is appropriate here, otherwise please move the post.

 

[zorro]

Yes your solution is correct.
The force due to pressure difference is ρgA(h-h1).
The additional force comes due to the impulse of water jet streaming out of the reservoir.