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
