Free Fall Q: Find Where Balls Cross Paths

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Here's the question: A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.

For the first ball, I said that the vo = 0, and by using the formula v^2 = vo^2 + 2ax, I found that the v = 21.7 m/s. I then applied this to the second ball, making its vo = 21.7 m/s. Here is where I became stuck.

I'm not sure how to find out where they cross paths - I figure I have to use simulataneous equations of some sort, but am not sure what variable I should solve for.
 
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Try to visualize the problem - how far did both of the balls travel until their paths crossed? This will give you an equation which you can solve for t and retrieve your solution.
 
I know that both of them will share the same t, but that still doesn't make any sense to me.
 
Of course they share the same time, but it's important that they 'swept' the length of 24m together. So, x1(t0) + x2(t0) = 24.
 
Ok, so I'm still pretty confused. I'm trying to work this out, and I've made this into a system of equations. Maybe you can take a look:

24-x = 1/2 (9.8)t^2 + 21.7t
x= 1/2(-21.7t) + 0

I know this is wrong, so I'm just confused where in my equations I'm wrong. Thanks again.
 
There is no x in your equation, only the time, t. Write down the equation of displacement for every ball, and then use the equation x1(t) + x2(t) = 24 (where x1(t) and x2(t) are the equations of motion for each ball, separately; one contains -g, and the other g). x mustn't confuse you; x is only the name of a function that depends on the time variable, t. We could have called it Z(t), or p(t), it really does not matter. x(t) is just a more conventional form.