a block of mass 77kg is placed at the top of a frictionless plane that is inclined 46degrees to the horizontal. The height of the plane is 23 meters. find the speed of the block at the bottom of the plane.
Show the steps please,
-thanks
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*23 = 450.8
V = 21.2 m/s.