x1 = .5 (9.89).52^2 = 1.32
x2 = .5 (9.8).27^2 = .36
Distance between is .96 and center of mass 1/3 distance from stone 2 = .31.
COM is therefore .36 + .31 = .67
b. v^2 = 2ax = 2(9.8).67 = 13.1
a. A stone is dropped at t = 0 s. A second stone, with a mass 2.0 times that of the first, is dropped from the same point at t = 0.25 s. How far from the release point is the center of mass of the two stones at t = 0.52 s? Assume that neither stone has yet reached the ground.
b. What is the speed of the center of mass of the two stone system at that time?
2 answers
I do not know