physics
posted by jerf on .
a ladder AP of length 5m inclined to a vertical wall is slipping over a horizontal surface with velocity 2m/s , when A is at a distance 3m from ground . what is the velocity of centre of mass at this moment?

Draw the diagram.
label the base of the ladder A, the top P, the corner O, and the cg C.
OA=3, AP=5, and you can compute OP as 4
OA^2+OP^2=AB^2
2 OA doa/dt +2 OP dop/dt=0 by taking the derivative (AB is a constant)
well, you know OA, OP, doa/dt=2, you can find dop/dt. At that instant, the center C has a velocity (1/2 doa/dt horizontal, and 1/2 dop/dt vertical)
so you have to add those two components as a vector