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Problem: A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free-fall, the ball has a centripetal acceleration of magnitude 2g. How far from the bottom of the chute does the ball land?

FREE Expert Solution

Centripetal acceleration:

α=v2R

Rearranging:

v = sqrt (αR)

v = sqrt (2gR)

Determining the time of free fall:

y (t) = y0 + v0yt + (1/2)ayt2

y0 = 2R, v0 = 0 m/s, and ay = -g

Therefore, y = 2R - (1/2)gt2

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Problem Details

A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free-fall, the ball has a centripetal acceleration of magnitude 2g. How far from the bottom of the chute does the ball land?

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