A 0.174-kg softball is pitched horizontally at 26.0 m/s. The ball moves in the opposite direction at 38.0 m/s after it is hit by the bat. A. What is the change in momentum of the ball? B. What is the impulse delivered by the bat? C. If the bat and ball are in contact for 0.80 ms, what is the average force the bat exerts on the ball?

The change in the momentum of the softball is 11.136 kgm/s.

The impulse delivered by the bat is 11.136 kgm/s.

The average force the bat exerts on the ball at the given time is 13,920 N.

The given parameters:

The change in the momentum of the softball is calculated as follows;

The impulse delivered by the bat is equal to change in momentum of the ball = 11.136 kgm/s.

The average force the bat exerts on the ball at the given time is calculated as follows;

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A)

B)

C) N

Explanation:

A )

Change in momentum is equal to Final momentum - Initial momentum

Final momentum

Initial Momentum

Thus, Change in momentum

B) Impulse delivered by bat is equal to the change in momentum i.e

C) Force is equal to change in momentum divided by time

Force

Force exerted by bat on ball is N

are there options? cause you ain’t. real specific

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer