1. The problem statement, all variables and given/known data A think cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, w/o slipping, from the top of an inclined plane that is 4.9m above the ground. When the first object reaches the bottom, what is the height above the ground of the other object? 2. Relevant equations (delta)x=.5(Vi+Vf)t 3. The attempt at a solution The problem also asked for the final linear velocity of both cylinders. Which I found to be: shell: 6.92965m/s solid:8m/s I tried to use the delta x equation above, but time is not given so I had to use ratio to cancel out the time: z: how far down the shell is solid: .5(8)t=(delta)x =4t shell: .5(6.93)t=(delta)x =3.465t 3.465t/4t = z/4.9 z=4.2445 answer: 4.9-4.2225=.655 however, this is wrong. Did I miss something?