Question

A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact time between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.

Answer 1

The average force between the ball and bat during contact is 3006.72 N.

Step-by-step explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the ball, u = 27 m/s

Time of contact between bat and ball is 2.5 ms,
`t=2.5* 10^(-3)\ s`

After striking the bat, it pops straight up to a height of 31.5 m. The final velocity of the ball is given by using third equation of motion as :

`v^2-u^2=2as`

a = -g

And initially, u = 0

`v=√(2gs)`

`v=√(2* 9.8* 31.5)`

v = -24.84 m/s (as it pops straight up)

Let F is the average force between the ball and bat during contact. It is given by :

`F=(m(v-u))/(t)`

`F=(0.145* (24.84 -(-27)))/(2.5* 10^(-3))`

F = 3006.72 N

So, the average force between the ball and bat during contact is 3006.72 N. Hence, this is the required solution.