# Phyiscs 211, two balls thrown

1. Jan 13, 2013

### n_nason

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

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

3. The attempt at a solution
I set the equations equal to eachother 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?

2. Jan 13, 2013

### cepheid

Staff Emeritus
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.

3. Jan 13, 2013

### 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?

4. Jan 13, 2013

### cepheid

Staff Emeritus
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.

5. Jan 13, 2013

### 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