Time for Two Balls to Reach Same Height

[n_nason]

Homework Statement

A red ball is thrown down with an initial speed of 1.0 m/s from a height of 27.0 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.8 m/s, from a height of 0.8 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 red ball is thrown, will the two balls be in the air at the same height.

Homework Equations

X=X(initial)+v(initial)t+1/2at^2

The Attempt at a Solution

I set the equations equal to each other for each ball (making x the same) so I got..
27+1t+4.9t^2=24.8(t-.4)+4.9(t-.4)^2
After factoring and so on, I got to..
35.96=19.88t
t=1.8
This is not correct, where am I going wrong?

 

[cepheid]

Welcome to PF n_nason!You're not being consistent with signs. Choose upward to be either the positive x-direction, or the negative x direction, but stick with your choice.

 

[n_nason]

So say I make the second half of my equation...
-24.8(t-.4)-4.9(t-.4)^2
Then my equation would be...
1t+4.9t^2+27=-24.8(t-.4)-4.9(t-.4)^2
My acceleration would not cancel, is that the right track?

 

[cepheid]

What happened to the 0.8 metres?

EDIT: also, the acceleration is in the SAME direction for both balls (downward). So why do the two 4.9s have different signs? Gravity didn't suddenly become upward.

Also, if your origin (x = 0) is at the ground level, and you've chosen upward to be the negative x-direction, then all of your positions should be negative as well.

 

[n_nason]

Oh, right that makes sense.
Still not getting the correct answer,
1t+4.9t^2+27=-24.8t+9.92+4.9t^2-3.92t+.784+.8
is my simplified answer
canceling acceleration and solving for t I get,
29.72t=-15.496
t=-1.92

 

[cepheid]

Again, you chose upward to be the negative x-direction, so all of your positions should be negative.