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?

Question

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(T-1)+.5(G)(T-1)^2

12m/s(T)+.5(G)T^2=18m/s(T-1)+.5(G)(T-1)
^2

I Cannt seem to get the right anwser of T=1.45s. What am I doing wrong?

Answer 1

You have it set up correcty, except for signs.

12t-4.9t^2=18(t-1)-4.9(t-1)^2
12t-4.9t^2=18t-18-4.9(t^2-2t+1)

now combine terms, and solve.

Answer 2

Thank You not sure how i did catch that lol.