a 1-kilogram block slides down a frictionless inclined plane with an angle of 35 degrees and 1.0 meters high. what is its velocity at the bottom of the plane?

Answer 1

V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*1 = 19.6
V = 4.43 m/s.

Answer 2

2.03