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A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseball travels a horizontal distance of 17.4 m and rotates through an angle of 44.1 rad. The baseball has a radius of 3.67 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_t=46.4532\ m.s^{-1}[/tex]

Explanation:

Given:

  • Linear horizontal speed of the ball, [tex]v_x=42.5\ m.s^{-1}[/tex]
  • distance travelled by the ball, [tex]x=17.4\ m[/tex]
  • angle rotated by the ball, [tex]\theta=44.1\ rad[/tex]
  • radius of the ball, [tex]r=3.67\ cm=0.0367\ m[/tex]

the time for which the ball travels:

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

[tex]t=\frac{17.4}{42.5}[/tex]

[tex]t=0.4094\ s[/tex]

Angular speed of the ball:

[tex]\omega=\frac{\theta}{t}[/tex]

[tex]\omega=\frac{44.1}{0.4094}[/tex]

[tex]\omega=107.7155\ rad.s^{-1}[/tex]

Now the tangential speed at the equator of the ball:

[tex]v_t=r.\omega+v_x[/tex] (the tangential speed due to rotation and the linear speed are having the same direction)

[tex]v_t=0.0367\times 107.7155+42.5[/tex]

[tex]v_t=46.4532\ m.s^{-1}[/tex]

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