Kinematics Ball throwing problem. Help

[mattyc33]

Homework Statement

A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.6 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.8 m/s from a height of 28.9 meters above the ground. The force of gravity due to the Earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

How long after the blue ball is thrown are the two balls in the air at the same height?

Homework Equations

v=vo+at
d=vot+1/2at^2
d=vt-1/2at^2
v^2=vo^2+2ad
d=t((vo+v)/2)

The Attempt at a Solution

Since yblue(t)=yredt
and
yred(t)=28.9+vred(t−2.8)+12a(t−2.8)2
yblue(t)=0.6+vbluet+12at2

So I tried to solve for t, but I never got the correct answer =(
So frustrating. If someone could help or even just give me a numerical answer so I can see where I went wrong that would be awesome!
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[gneill]

What values are you assigning to your variables yred, vred, a, yblue, vblue? Signs matter.
Can you show more of your attempt to solve for t?

 

[mattyc33]

I did not assign my yred nor yblue to anything because yred=yblue. So I just plugged in what yred and yblue are equaled to and I made them equal each other so that I had two formulas which only had one variable in it (t).

I assigned my vred as (t(6.8+15.0404))/2 which made it even more complicated. I found the velocity in an earlier equation but I don't even know if that is right =(.

I assigned my vblue as 23m/s... which is initial.

I assign a to 9.81 (gravity constant).

After doing this question for what seems like two hours my head is filled with confusion and it's hard to think. When I solve for t with all this information I get approximatly 1.09 seconds which is incorrect.

 

[gneill]

Well, your approach is okay, but I can't tell if the values that you are plugging into the variables are consistent with their signs in the equations that you've written.

Suppose we set up the following variables and constants (all constants will have positive values -- we'll deal with directions in the operational signs in the equations we write):

Initial upward speed of blue ball  :  vb = 23 m/s
Initial height of blue ball        :  hb = 0.6 m
Initial downward speed of red ball :  vr = 6.8 m/s
Initial height of red ball         :  hr = 28.9m
Position of blue ball at some time :  pb(t)
Position of red ball at some time  :  pr(t)
Time delay in launch of red ball   :  Δt = 2.8 s
Gravitational acceleration         :  g = 9.81 m/s/s

Using these variables and constants to rewrite your equations:
pb(t) = hb + vb\;t - \frac{1}{2} g\;t^2
pr(t) = hr - vr\;(t - \Delta t) - \frac{1}{2} g\;(t - \Delta t)^2
Now you should be able to equate these two in search of the time when the heights are the same. Can you plug in the known numerical values and simplify to a quadratic in t?

 

[mattyc33]

It worked! Thanks =)