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?

Answer 1

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