A cylindrical can of paint with (I = .5MR^2) starts from rest and rolls down a roof as shown in the link http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1424481748 Determine how far the can lands from the edge of the house. This is my work, but I am not getting the correct answer. Please tell me where I messed up. The correct answer is 6.18m mgH = (1/2)MV_cm^2 (1+B) gH = (1/2)V_cm^2 (1+B) 3g = (1/2)V_cm^2(1+.5) 3g = (3/4)v_cm^2 4g = v_cm^2 v = 2SQRT(g) Now, I know that x = x_0 + v_o*cos(theta)*t x = 2SQRT(g)*cos(30)*t To solve for t, I used y y = 10 - 2SQRT(g)*sin(30)*t - (1/2)gt^2 0 = 20 - 4SQRT(g)*sin(30)*t - gt^2 0 = gt^2 + 2SQRT(g)*t - 20 t = 2SQRT(g) +/- SQRT[4g-4(g)(-20)]/2g t = 7.72 seconds or 4.79 x = 2SQRT(g)*cos(30)*7.72 x = 41.8 x = 2SQRt(g)*cos(30)*4.79 x = 26.0 The correct answer is 6.18. My answer is too far off, but I don't know where I went wrong.