# Relative motion (2 arrow shot up find meeting distance)

1. Jan 26, 2010

### GregR

1. The problem statement, all variables and given/known data

Two arrows are fired straight up. The first is at 40 m/s and the second at 60 m/s and 3 seconds after arrow one. At what hieght will the arrows meet in the air.

2. Relevant equations
?
s=ut+1/2 at^2

3. The attempt at a solution

s1=s2
t2=t1+3
u1(t2+3)+1/2a(t2+3)^2=u2t2+1/2at2^2

2. Jan 26, 2010

### rl.bhat

Hi GregR, welcome to PF.
Your approach is correct.
But
t2=t1+3 it should be t1=t2+3.
In the equation you have substituted correctly.
Now proceed.

3. Jan 26, 2010

### GregR

So then:
u1t2+3u1+.5at22+4.5a=u2t2+.5at22

u1=40m/s
u2=60m/s
a=-9.81m/s2

The .5at22 on each side should cancel each other out leaving:

u1t2+3u1+4.5a=u2t2

Solving:

40t2+3(40)+4.5(-9.81)=60t2
40t2+120-44.145=60t2
20t2=75.855
t2=75.855/20
t2=3.79275 seconds

4. Jan 26, 2010

### GregR

If I use this as the time and recalculate the distance for each arrow it doesn't jive.

Through trial and error I found the meeting time to be 1.5345 seconds at a distance of 80.52 meters from the ground.

Not sure if I'm missing something or its just a minor error on my part.

5. Jan 26, 2010

### rl.bhat

Your calculation is wrong.
The equation will be
u1(t+3) - 0.5*g*(t+3)^2 = u2*t - 0.5*g*t^2
Simplify the equation and solve for t.

6. Jan 26, 2010

### GregR

Using this formula:
u1(t+3) - 0.5*g*(t+3)^2 = u2*t - 0.5*g*t^2

From this I simplified as follows:

u1t +3u1-0.5*9.81*(t2+9) = u2*t - 0.5*9.81*t2

u1t +3u1-4.905*(t2+9) = u2*t - 4.905*t2

u1t +3u1-4.905*t2-4.905*9 = u2*t - 4.905*t2

The -4.905*t2 on both sides should cancel out leaving.

u1t +3u1-44.145= u2*t

Inserting u1=40m/s and
u2=60m/s

40*t + 30*40 - 44.145 = 60*t
40*t + 120 - 44.145 = 60*t
75.855 = 60*t - 40*t
20*t = 75.855
t = 75.855 / 20
t = 3.79275

I wrote it up step by step and got the same results.
Not sure where I'm going wrong here.
Please let me know.
Thanks

7. Jan 27, 2010

### rl.bhat

(t+3)^2 = t^2 + 6*t + 9