Question

A boy whirls a stone in a horizontal circle of radius 1.1 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? m/s2

Answer 1

`212.8 m/s^(2)`

Step-by-step explanation:

Time taken by stone to cover horizontal distance

`t=\sqrt{\frac {2h}{g}}` where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

`t=\sqrt{\frac {2*2.1}{9.81}}`= 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

`a=\frac {v^(2)}{r}` where r is radius of circle, substituting r with 1.1m

`a=\frac {15.3^(2)}{1.1}`

`a=212.8 m/s^(2)`

Therefore, centripetal acceleration is
`212.8 m/s^(2)`