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 teh 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.