Answer:
a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
            Em₀ = U = mg h
final point. Lowest point
             = K
 = K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
            K = ½ m  + ½
 + ½  w²
 w²
angular and linear speed are related
            v = w r
            w = v / r
             K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
             Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
              Em₀ = Em_{f}
              mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)
              h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
              I_{cm} = ⅔ m r²
we substitute
             h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
             h = ½ v_{cm }^{2}/g    5/3
              h = 5/6 v_{cm }^{2} / g
            
let's calculate
            h = 5/6 6.1 2 / 9.8
            h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
           v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
               I_{cm} = ½ m r²
we substitute
              v_{cm } = √ [2gh / (1 + ½)]
              v_{cm } = √(4/3 gh)
let's calculate
              v_{cm } = √ (4/3 9.8 3.16)
              v_{cm }^ = 6.43 m / s