If a ball is rolled down a frictionless inclined plane from a height of 3 feet, to what height would it roll up to on the other side?

It will roll up to its initial height from which it rolled from on the first hill

Explanation:

The given information are;

The height from which the ball is rolled = 3 feet[

The equation of the potential energy of the ball as it rolled down the frictionless inclined plane is given by the following equation;

m × g × h₁ = 1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r]

Where;

m = The mass of the ball

h = The from which the ball is rolled down

v = The velocity of the ball

r = The radius of the ball

Therefore, all the potential energy, m × g × h, is converted into kinetic energy when the ball reached the ground

Similarly, as the ball rolls up on the other side, by the principle of energy conservation, all the kinetic energy is transformed into potential energy as follows

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₂, therefore, we have;

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₁ =  m × g × h₂

Therefore since m × g × h₁ =  m × g × h₂, then h₁ = h₂, the ball rises to the same height it started rolling from on the first hill.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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