Question

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

Answer 1

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, $r=3.67\ cm=0.0367\ m$

the time for which the ball travels:

$t=\frac{x}{v_x}$

$t=\frac{17.4}{42.5}$

$t=0.4094\ s$

Angular speed of the ball:

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

$\omega=\frac{44.1}{0.4094}$

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

Now the tangential speed at the equator of the ball:

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

$v_t=0.0367\times 107.7155+42.5$

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