Physics
posted by Kayla .
You roll a 5 kg bowling ball (R=20cm) with an initial speed of 10 m/s up a 30 degree ramp that is 12 m long. Will the ball reach the top of the ramp? (support your answer)

What is the translational (1/2 mv^2) PLUS the initial rotational energy (1/2 I w^2, where w=vr). Will that equal mgh?

Thanks bobpursley! I was somewhere close to that idea. I had mgh=1/2mv^2 + 1/5mv^2. I let h=12sin30. I was solving for velocity for some reason, doh!

so I have mgh = 300.
i have my rotational energy = 350.
so it will reach the top, correct? 
Try to obtain the general solution of the problem, i.e.,
KE=PE
KE(tr) + KE(rot) =PE
0.7m•v²/2 = m•g•h
h=0.7•v²/g=70/9.8=7.14 m.
h(real)=s•sinα=12•0.5=6 m
Since 7,14>6, the ball‘ll
rich the top of the ramp.