Question

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?

Answer 1

Answer:

$v = 3.61 m/s$

Explanation:

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

so we will have

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

$t = \frac{24.7}{49.4}$

$t = 0.5 s$

now in the same time ball is turned by angle

$\theta = 52.7 rad$

now we know that

$\theta = \omega t$

$52.7 = \omega (0.5)$

$\omega = 105.4 rad/s$

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

$v = r\omega$

$v = 0.0343(105.4)$

$v = 3.61 m/s$