A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 500 ms. (a) How far below the release point is the center of mass of the two stones at t = 850 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time? I started by solving for how far they fell. I'll use m1 for the first stone dropped and 2m1 for the second stone dropped, y1 for how far the first stone dropped and y2 for how far the second stone dropped. y = .5gt2 y1 = .5(9.8)(0.850)2 y1 = 3.54 m y2 = .5(9.8)(0.350)2 y2 = 0.60 m For part a I got the equation ycom = (m1(3.54) + 2m1(0.60))/3m1 But how do I solve this when I'm not given the value of m? I'm not sure how to solve part b. Is it the same idea except I replace y with v? I'm saying vcom = (m1v1 + 2m1v2) / 3m1 Help please.