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A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 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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Answer:
233.23 m/s^2

Step-by-step explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground


1.5=0+(gt^2)/(2)

t=0.55 s

Initial horizontal velocity at the time of break is given by u


R=u* t


10=u* 0.55

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by


a_c=(u^2)/(r)=(18.07^2)/(1.4)=233.23 m/s^2

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