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**Physics Kinematics Homework Help!!?**

"While standing on a bridge 18.3 m above the ground, you drop a stone from rest. When the stone has fallen 2.50 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction."

After finding the velocity of the first stone (AFTER falling 2.50 m) to be -7m/s, I tried to solve for initial velocity of the second stone by setting the times equal to each other (displacement for the first stone being 15.8, and for the second stone 18.3).

I used the equation x = v0*t + (1/2)*a*t2

and of course I got an ugly quadratic-equationey mess trying to get t on one side. There has to be an easier way to do this...