Question

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

Answer 1

a = 162.7 m/s²

Step-by-step explanation:

given,

radius = 1.5 m

height above ground level = 2 m

horizontal distance = 10 m

magnitude of centripetal acceleration = ?

we know,

centripetal acceleration

a =
`(v^2)/(r)`

using equation of motion

`s = u t + (1)/(2)at^2`

`s =(1)/(2)gt^2`

`t = \sqrt{(2s)/(g)}`

`t = \sqrt{(2* 2)/(9.8)}`

t = 0.64 s

velocity

`v = (s)/(t)`

`v = (10)/(0.64)`

v = 15.625 m/s

centripetal acceleration

a =
`(15.625^2)/(1.5)`

a = 162.7 m/s²