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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?

Respuesta :

Answer:

[tex]a_c = 120.4 m/s^2[/tex]

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

[tex]H = \frac{1}[2}gt^2[/tex]

now we have

[tex]1.9 = \frac{1}{2}(9.81)t^2[/tex]

now we have

[tex]t = 0.622 s[/tex]

now the speed of the stone is given as

[tex]v = \frac{x}{t}[/tex]

[tex]v = \frac{8.9}{0.622} = 14.3 m/s[/tex]

now for finding centripetal acceleration we have

[tex]a_c = \frac{v^2}{R}[/tex]

we know that radius of circle is

R = 1.7 m

now we have

[tex]a_c = \frac{14.3^2}{1.7}[/tex]

[tex]a_c = 120.4 m/s^2[/tex]

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