Starting from rest, a basketball rolls from the top to thebottom of a hill, reac

Question

Starting from rest, a basketball rolls from the top to thebottom of a hill, reaching a translational speed of 6.6 m/s. Ignore frictional losses. (a) What is theheight of the hill? (b) Released from restat the same height, a can of frozen juice rolls to the bottom ofthe same hill. What is the translational speed of the frozenjuice can when it reaches the bottom? I'm having trouble with this question because they don'tprovide the inertia or the radius of the sphere (basketball) or thecylinder (frozen juice can). I got an answer of 2.22 m forpart a, but I used the conservation of energy without accountingfor the rotational energy, because like I said, they don't providethe inertia or radius. Any help would be greatlyappreciated. Starting from rest, a basketball rolls from the top to thebottom of a hill, reaching a translational speed of 6.6 m/s. Ignore frictional losses. (a) What is theheight of the hill? (b) Released from restat the same height, a can of frozen juice rolls to the bottom ofthe same hill. What is the translational speed of the frozenjuice can when it reaches the bottom? I'm having trouble with this question because they don'tprovide the inertia or the radius of the sphere (basketball) or thecylinder (frozen juice can). I got an answer of 2.22 m forpart a, but I used the conservation of energy without accountingfor the rotational energy, because like I said, they don't providethe inertia or radius. Any help would be greatlyappreciated.

Explanation / Answer

.    Here   mgh =0.5mv2   .
   or gh= 0.5 v2   .
   or    h = (0.5 *(6.6)2)/9.8 = 2.222 m
.
b) As the object is releasing from the sameheight .    m g h = 0.5 m v2   .
      v = sqrt(2 g h) = sqrt(2 * 9.8 *2.222) = 6.599 m/s . Hope this helps u!

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