A blue ball is thrown upward with an initial speed of 23.9 m/s, from a height of 0.7 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.6 m/s from a height of 31.1 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? I have checkedsig-figs, redone the equation, used wolframalpha, and yet the question is wrong. Thanks for the help.

Question

jt950 · Answer

I get the time t=0.29734 but the time -2.9second for red ball to be thrown will be negative 
so i think the question might be wrong

anonymous · Answer

That is what I thought, but I did some research and even asked my teacher and he said that there was an answer to this question. I tried using the quadratic equation and setting the respective equations equal to each other, but they come out wrong.

goformit100 · Answer

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anonymous · Answer

$$0.7+23.9(t+2.9)-\frac{ 1 }{ 2}g( t+2.9)^2=31.1-9.6t-\frac{ 1 }{ 2 }g t^2$$Solving this equation I get: t=0.46 sec, then the time since the blue ball is launched is:t=0.46+2.9=3.36 sec

jt950 · Answer

Sorry I was mistaken that red ball to be thrown "up" 
LOL

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