Question

A 3.0 kg block slides along a horizontal frictionless surface with a speed of 9.8 m/s. The block makes a smooth transition to a frictionless ramp inclined at an angle of 35.0

Answer 1

mv^2/2 = mgh
height h is opposite to the angle 35 deg.
distance moved along the plane before coming to rest is x
Then ,
h = v^2/2g = 96.04/2*9.8 = 4.9 m
sin(35) = h/x
x = h/sin(35) = 8.54288 m

Answer 2

mv^2/2 = mgh is the correct relation
The height h is opposite to the angle 40deg.
The distance moved along the plane before coming to rest is x
Then , h = v^2/2g = 49/2*9.8 = 2.5 m
sin40 = h/x
x = h/sin40 = 3.89 m