Starting from rest, a basketball rolls from top of a hill to the bottom, reaching a translational speed of 6.8 m/s. ignore frictional force losses. a. what is the height of the hill? b. released from the same height, a can of frozen juice rolls to the bottom of the same hill. c. what is the translational speed of the frozen juice can when it reaches the bottom?

Explanation:

for baseball

(a) Let the mass of the baseball is m.

radius of baseball is r.

Total kinetic energy of the baseball, T = rotational kinetic energy + translational kinetic energy

T = 0.5 Iω² + 0.5 mv²

Where, I be the moment of inertia and ω be the angular speed.

ω = v/r

T = 0.5 x 2/3 mr² x v²/r² + 0.5 mv²

T = 0.83 mv²

According to the conservation of energy, the total kinetic energy at the bottom is equal to the total potential energy at the top.

m g h = 0.83 mv²

where, h be the height of the top of the hill.

9.8 x h = 0.83 x 6.8 x 6.8

h = 3.93 m

(b) Let the velocity of juice can is v'.

moment of inertia of the juice can = 1/2mr²

So, total kinetic energy

T = 0.5 x I x ω² + 0.5 mv²

T = 0.5 x 0.5 x m x r² x v²/r² + 0.5 mv²

m g h = 0.75 mv²

9.8 x 3.93 = 0.75 v²

v = 7.2 m/s

neither accurate nor precise.