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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 ?

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

1 Answer

Explanation:

The centripetal acceleration is given by:

#a=v^2/r#

We can find #v# from the information given. Smart converter pro 3 0 1 0 download.

Considering the vertical component of the motion only we can say: https://bestffiles361.weebly.com/windows-auf-macbook-pro.html. Photograv crack.

#s=1/2'g't^2#

Apple help ios 11. #:.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'# https://downcfile403.weebly.com/fallout-4-sniper-outfit-mod.html.

Family tree maker 23 3 0 570 cc. So now we can get the centripetal acceleration:

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

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