Sphere rolling uphill a hollow sphere is rolling along a horizontal floor at 5.0 m/s when it comes to a 30 degree incline. how far up the incline does it roll before reversing direction?

The sphere has translational and rotational kinetic energy. 
Trans equation: Ek = ½mv² 
Rotation equation: Ek = ½Iω² 
substitute I = 2mr²/3 
and ω = v / r: 
rotat Ek = ½(2mr²/3)(v/r)² = (1/3)mv² 
Then total E = (½ + 1/3) mv²
= (5/6)mv²
= (5/6)m(5m/s)²
= m*20.8m²/s² 
The sphere attains a vertical height such that this E is converted into potential E: 
mgh = m*9.8m/s²*h = m*20.8m²/s² 
mass m cancels: 
= 20.8 / 9.8
so h = 2.12 m
This translates into a distance along the incline of s = 2.12 m / sin30º = 4.24 m