Phyics
posted by Joshua .
A Rock Is Thrown Vertically upward with a speed of 12m/s. Exactly 1.00s later a ball is thrown up vertically along the same path with a speed of 18m/s. At what time will they strike each other?
so far I got this but it does not seem to be working out for me.
G=9.80m/S^2
Y1=0+12m/s(T)+.5(G)T^2
Y2=0+18m/s(T1)+.5(G)(T1)^2
12m/s(T)+.5(G)T^2=18m/s(T1)+.5(G)(T1)
^2
I Cannt seem to get the right anwser of T=1.45s. What am I doing wrong?

You have it set up correcty, except for signs.
12t4.9t^2=18(t1)4.9(t1)^2
12t4.9t^2=18t184.9(t^22t+1)
now combine terms, and solve. 
Thank You not sure how i did catch that lol.
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