Constant acceleration motion problem,

[chroncile]

Homework Statement

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?

Homework Equations

d = vf * t - 0.5(a)(t)²
t_A1 = t_A2 + 2

The Attempt at a Solution

d_A1 = 25t - 4.9t²
d_A2 = 25t - 4.9t²

Since d_A1 = d_A2

25t_A1 - 4.9t_A1² = 25t_A2 - 4.9t_A2²

Since t_A1 = t_A2 + 2

25(t_A2 + 2) - 4.9(t_A2 + 2)² = 25t_A2 - 4.9t_A2²

19.6t = 50-19.6
t = 1.55 s

The correct answer is 3.55 s

 

[berkeman]

Homework Statement

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?

Homework Equations

d = vf * t - 0.5(a)(t)²
t_A1 = t_A2 + 2

The Attempt at a Solution

d_A1 = 25t - 4.9t²
d_A2 = 25t - 4.9t²

Since d_A1 = d_A2

25t_A1 - 4.9t_A1² = 25t_A2 - 4.9t_A2²

Since t_A1 = t_A2 + 2

25(t_A2 + 2) - 4.9(t_A2 + 2)² = 25t_A2 - 4.9t_A2²

19.6t = 50-19.6
t = 1.55 s

The correct answer is 3.55 s

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?

 

[berkeman]

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?

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.

 

[chroncile]

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

 

[berkeman]

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

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

 

[chroncile]

Can you please explain that?

 

[berkeman]

Can you please explain that?

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.

 

[chroncile]

Okay, I get it now, thanks

 

[berkeman]

Okay, I get it now, thanks

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.