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? (Your answer for the distance the ball travels from the end of the chute should contain R.)

Because, before free fall if velocity of ball is v (horizontal)
v^2/R = 2*g
=> v = sqrt(2*g*R)

In ...

Solution Summary

We start this solution by solving for the velocity of ball before free fall in the horizontal direction: v^2/R = 2*g => v = sqrt(2*g*R). Full calculations and answer for distance travelled from chute are included.

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