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 16.0 m and rotates through an angle of 44.5 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:

The tangential speed of a point on the "equator" of the baseball is 4.33 m/s.

Explanation:

Given that,

Linear speed of base ball = 42.5 m/s

Distance = 16.0 m

Angle = 44.5 rad

Radius of baseball = 3.67 cm

We need to calculate the flight time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{16.0}{42.5}$

$t=0.376\ sec$

We need to calculate the number of rotation

Using formula of number of rotation

$n=\theta\time 2\pi$

$n=\dfrac{44.5}{2\pi}$

$n=7.08$

We need to calculate the time for one rotation

Using formula of time

$T=\dfrac{t}{n}$

Put the value into the formula

$T=\dfrac{0.376}{7.08}$

$T=0.053\ sec$

We need to calculate the circumference

Using formula of circumference

$C=2\pi\times r$

Put the value into the formula

$C=2\pi\times3.67\times10^{-2}$

$C=0.23\ m$

The tangential speed is equal to the circumference divided by the time. it takes to complete one rotation.

We need to calculate the tangential speed

Using formula of tangential speed

$v=\dfrac{C}{T}$

Put the value into the formula

$v=\dfrac{0.23}{0.053}$

$v=4.33\ m/s$

Hence, The tangential speed of a point on the "equator" of the baseball is 4.33 m/s.