1. The problem statement, all variables and given/known data You throw a stone straight up with initial V of 24.5m/s and 2.0 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? 2. Relevant equations I'm not sure if I'm doing the question correctly. My attempt at it is below. 3. The attempt at a solution 1st Stone d = Vt d = 24.5t 2nd Stone d = 24.5t + .5(9.8)t^2 after 2 seconds, 24.5 x 2 = 49, so d + 49 = 24.5t +.5(9.8)t^2 I then plugged in the d equation from the 1st stone 24.5t + 49 = 24.5t + 4.9t^2 4.9t^2 = 49 t = 3.162 seconds is when the two stones are at the same distance. Then i assume that you just plug in the t-value in the distance equation for the 1st stone but i keep getting an incorrect answer. Any help please?