Posted by **Sam** on Saturday, July 10, 2010 at 5:57pm.

A boy whirls a stone in a horizontal circle of radius 1.9 m and at height 2.1 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?

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**drwls**, Sunday, July 11, 2010 at 2:44am
The centripetal acceleration was V^2/R, where R = 1.9 m.

You need to compute V, the speed when in circular motion. First compute the time T that it takes the stone to fall 1.9 m.

1.9 m = (1/2) g T^2

V = Vx, the horizontal component of velocity after the string breaks

V = (10 m)/T

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**Anonymous**, Thursday, September 22, 2011 at 12:39am
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