Question

A boy whirls a stone in a horizontal circle of radius 1.7 m and at height 1.9 m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 8.9 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?

Answer 1

`a_c = 120.4 m/s^2`

Step-by-step explanation:

As we know that stone is revolving in horizontal circle

so here height of the stone is H = 1.9 m

now by kinematics we can find the time to reach it at the ground

`H = \frac{1}[2}gt^2`

now we have

`1.9 = (1)/(2)(9.81)t^2`

now we have

`t = 0.622 s`

now the speed of the stone is given as

`v = (x)/(t)`

`v = (8.9)/(0.622) = 14.3 m/s`

now for finding centripetal acceleration we have

`a_c = (v^2)/(R)`

we know that radius of circle is

R = 1.7 m

now we have

`a_c = (14.3^2)/(1.7)`

`a_c = 120.4 m/s^2`