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lord [1]
3 years ago
12

A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba

ll 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
Physics
1 answer:
Xelga [282]3 years ago
4 0

Answer:

v_t=46.4532\ m.s^{-1}

Explanation:

Given:

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

<u>the time for which the ball travels:</u>

t=\frac{x}{v_x}

t=\frac{17.4}{42.5}

t=0.4094\ s

<u>Angular speed of the ball:</u>

\omega=\frac{\theta}{t}

\omega=\frac{44.1}{0.4094}

\omega=107.7155\ rad.s^{-1}

<u>Now the tangential speed at the equator of the ball:</u>

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

v_t=0.0367\times 107.7155+42.5

v_t=46.4532\ m.s^{-1}

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