A 6.0 kg block,starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the incline its speed is vf. At what distance from the top of the incline is the speed of the block .5 vf

Answer 1

Look at the energy equation:

PElost=kinetic energy changed.
mgh=1/2 m v^2

v= sqrt 2gh
so when v halves, h must be (sqrt 2)/2