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## Homework Statement

A blue ball is thrown upward with an initial speed of 20.4 m/s, from a height of 0.5 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.8 m/s from a height of 23.9 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

Xf=Xi+Vit+0.5a^2

## The Attempt at a Solution

Blue ball:

Vi=20.4 m/s

a=-9.81

Xi=0

t=t-2.5

Red ball:

Vi=-9.8

a=-9.81

xi=23.4

t=t

The height is different because i chose the blue ball as reference point.

I plugged the values into the equation and equated them giving me:

20.4(t-2.5)-4.9(t+2.5)^2=23.4-9.8t-4.9t^2

further simplifying i get....

-4.1t-4.9t^2-81.6=23.4-9.8t-4.9t^2

then...

5.7t=105

t=18.4s

which is wrong. can someone tell me where i went wrong?

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