Centripetal acceleration:

$\overline{){\mathbf{\alpha}}{\mathbf{=}}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}}$

Rearranging:

v = sqrt (αR)

v = sqrt (2gR)

Determining the time of free fall:

y (t) = y_{0} + v_{0y}t + (1/2)a_{y}t^{2}

y_{0} = 2R, v_{0} = 0 m/s, and a_{y} = -g

Therefore, y = 2R - (1/2)gt^{2}

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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