Hydroelectric Power: How Much Water for a 30m Head?

[Kevin2341]

Homework Statement

Assuming 100% efficient energy conversion, how much water
stored behind a hydroelectric plant with a head of 30m would
be required to charge the battery?

Homework Equations

?
MAYBE:
w=1000QgH j\s, where Q is the discharge rate @ m^3\s, 1000Q kg\s, H = net water head in m (30m), g is 9.8 m\s^2

The Attempt at a Solution

I don't even know where to begin with this question. My last question was a simple unit conversation question about how much energy was stored in a 12v 50Ah rated car battery, which I got 2,160,000 joules for my answer. I don't even know how to begin to tackle this question because the formula I gave was one I found on the internet that looked like it might apply.

What I'm thinking with that question however, is that perhaps I can use my figure from the total energy of the battery equaling:
2,160,000W = 1000*Q*9.8m\s^2*30m j (I borrowed the seconds unit from the j\s to turn my 2160000 joules into watts on the left hand side.)

2,160kW = 294Q kg\s m^2\s^2 j
7346.938 = Q kg m^2/s^3 j

The numbers look way off to me, and also the units have some serious funkiness going on with them which leads me to think this method isn't correct.

Any help?

 

[rude man]

1. How much energy is required to charge a dead 50Ahr battery (assumedly 12V)?
2. Potential energy of 1 kg of water 30m high is how many Joules? So how many of those Joules = 12V*50A*3600sec?