Relative motion (2 arrow shot up find meeting distance)

[GregR]

Homework Statement

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.

Homework Equations

?
s=ut+1/2 at^2

The Attempt at a Solution

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

 

[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.

 

[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

 

[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.

 

[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.

 

[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

 

[rl.bhat]

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