Starting from rest, a basketball rolls from the top of a hill tothe bottom, reac

Question

Starting from rest, a basketball rolls from the top of a hill tothe bottom, reaching a translational speed of 3.60 m/s. Ignore frictional losses. (a) What is the height of the hill?
m

(b) Released from rest at the same height, a can of frozen juicerolls to the bottom of the same hill. What is the translationalspeed of the frozen juice can when it reaches the bottom?
m/s (a) What is the height of the hill?
m

(b) Released from rest at the same height, a can of frozen juicerolls to the bottom of the same hill. What is the translationalspeed of the frozen juice can when it reaches the bottom?
m/s

Explanation / Answer

According to conservation of energy for a basketball
         m g h =(1/2)I2 + (1/2)mV2      I = (2/3)mR2 and =V /R         So         m g h = (5/6)mV2          h =5V2 / 6 g            =5 (3.60 m/s)2 / 6 (9.8)             = 1.102m For a can of juice then     M g h=  (1/2)I2 + (1/2) MV2    here I = (1/2)MR2 and = V /R        V = gh            = (9.8)(1.102m)            =3.286 m/s

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