posted by Genny .
A block of mass 77.0kg is placed at the top of a frictionless plane that is inclined 46degrees to the horizontal. The height of the plane is 23.0meters. Find the speed of the block and the bottom of the plane.
The correct answer is: 21.2m/s...
I've tried this equation every way I can think of and I just cannot seem to get it!! >.<
Thank you for your help!
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*23 = 450.8
V = 21.2 m/s.