1. Sep 21, 2010

### chroncile

1. The problem statement, all variables and given/known data
An arrow is shot vertically upward with an initial speed of 25 m/s. Two seconds later, another arrow is shot upward with the same speed. On its way up, the second arrow meets the first arrow on its way down. Assume that both arrows experience a constant downward acceleration of 9.8 m/s2.

How long does it take (from the time the first arrow is shot) for the arrows to meet?

2. Relevant equations
d = vf * t - 0.5(a)(t)2
tA1 = tA2 + 2

3. The attempt at a solution
dA1 = 25t - 4.9t2
dA2 = 25t - 4.9t2

Since dA1 = dA2

25tA1 - 4.9tA12 = 25tA2 - 4.9tA22

Since tA1 = tA2 + 2

25(tA2 + 2) - 4.9(tA2 + 2)2 = 25tA2 - 4.9tA22

19.6t = 50-19.6
t = 1.55 s

The correct answer is 3.55 s

2. Sep 21, 2010

### Staff: Mentor

I believe it's just that you have this equation off: tA1 = tA2 + 2

The 2nd arrow is shot 2 seconds after the first, not the other way around. Does that fix it?

3. Sep 21, 2010

### Staff: Mentor

Actually, the only error is related to that equation. You forgot to add the 2 seconds to the time of the 2nd arrow being shot. You solved tor t2, and they asked for t1.

4. Sep 21, 2010

### chroncile

I tried switching them around, but I still ended up with 1.55 s, help?

5. Sep 21, 2010

### Staff: Mentor

Did you see my last post about adding the 2 seconds?

6. Sep 21, 2010

### chroncile

Can you please explain that?

7. Sep 21, 2010

### Staff: Mentor

The solution you got is for the time from firing the 2nd arrow up. But you fired the 1st arrow 2 seconds earlier. Therefore the time for the arrows to meet from the time you fired the first arrow is 1.55s + 2.0s = 3.55s.

8. Sep 21, 2010

### chroncile

Okay, I get it now, thanks

9. Sep 21, 2010

### Staff: Mentor

You're welcome. You did a good job setting up and solving the problem. It makes it a lot easier to help when we get to see all of your work like that.