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

`v = 3.61 m/s`

Step-by-step 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 = (x)/(v_x)`

`t = (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`