# Free fall question

1. Sep 18, 2006

### tigerguy

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.

2. Sep 18, 2006

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.

3. Sep 18, 2006

### tigerguy

I know that both of them will share the same t, but that still doesn't make any sense to me.

4. Sep 18, 2006

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.

5. Sep 19, 2006

### tigerguy

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.

6. Sep 19, 2006