A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 10m ?

Question

What is the magnitude of the centripetal acceleration of the stone during the circular motion?

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Answer 1 · 2015-12-18T17:25:30

#162.7"m/s"^2#

The centripetal acceleration is given by:

#a=v^2/r#

We can find #v# from the information given.

Considering the vertical component of the motion only we can say:

#s=1/2"g"t^2#

#:.t=sqrt((2s)/g)#

#:.t=sqrt((2xx2)/(9.8))=0.64"s"#

The horizontal component of velocity shares the same time of flight and is constant so:

#v=10/0.64=15.62"m/s"#

So now we can get the centripetal acceleration:

#a=v^2/r=15.62^2/1.5=162.7"m/s"^2#