Suppose you throw a stone straight up with an initial velocity of 18.0 m/s and, 0.5 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?

Answer 1

height1=18t-1/2 g t^2
height2=18(t+.5)-1/2 g (t+.5)^2

set them equal, solve for time t. then calculate the height using either equation