Abowling ball encounters a 0.760 m vertical rise on the way back to the ball rack, as the drawing illustrates. ignore frictional losses and assume that the mass of the ball is distributed uniformly. the translational speed of the ball is 4.50 m/s at the bottom of the rise. find the translational speed at the top.

v = 2.31 m/s

Explanation:

Kinetic energy will be converted partially into potential energy and rest will remain in the form of kinetic energy .

Initial kinetic energy = 1/2 mu²

u is initial velocity at the bottom .

Potential energy at the top

= mgh

Kinetic energy at the top

= 1/2 m v²

According to conservation of energy

1/2 mu² = mgh + 1/2 m v²

1/2 u² = gh + 1/2  v²

.5 x 4.5² = 9.8 x .76 + .5 v²

10.125 = 7.448 + .5 x v²

v = 2.31 m/s

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it doesn't. it was imagined up in the middle ages.