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A baseball pitcher throws a baseball horizontally at a linear speed of 49.4 m/s. Before being caught, the baseball travels a horizontal distance of 24.7 m and rotates through an angle of 52.7 rad. The baseball has a radius of 3.43 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

Respuesta :

Answer:

[tex]v = 3.61 m/s[/tex]

Explanation:

As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s

so we will have

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

[tex]t = \frac{24.7}{49.4}[/tex]

[tex]t = 0.5 s[/tex]

now in the same time ball is turned by angle

[tex]\theta = 52.7 rad[/tex]

now we know that

[tex]\theta = \omega t[/tex]

[tex]52.7 = \omega (0.5)[/tex]

[tex]\omega = 105.4 rad/s[/tex]

now the tangential speed of a point at equator is given as

[tex]v = r\omega[/tex]

[tex]v = 0.0343(105.4)[/tex]

[tex]v = 3.61 m/s[/tex]

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