Translational Speed of a Bowling Ball

1. A bowling ball with a mass of 6.95 kg and a radius of 0.194 m starts from rest at a height of 2.21 m and rolls down a 49.2 degree slope. What is the translational speed of the ball when it leaves the incline?


2. Unsure! Here's what I've got, though:
ω = v/r
mgh=1/2mv²+1/2Iω²
And so: ω= √(2mgh/(mr²+I))
I of a solid sphere = 2/3(m)(r²)


3. The attempt at a solution
I = 2/3(6.95)(.194²) = 0.174
ω= √(2(6.95)(9.8)(2.21)/((6.95)(.194²+0.174)) = √(301.05)/(0.436) = 26.29
v = ωr
v = 26.20 X 0.194 = 5.1
Alas, 5.1 is not the correct answer. I'm sure I need to factor in the angle somewhere, but I don't know how! Help, please!

 

Oh my goodness, has this been my problem the whole time?

I tried it, it was! You are such a lifesaver, thank you so much! I couldn't for the life of me figure out what I was doing wrong! Thanks again!