A 0.140-kg baseball is thrown with a velocity of 27.1 m/sIt is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?

The bat and the ball were in contact for 1.8 x 10⁻³ s

Explanation:

Given;

mass of baseball, m = 0.14 kg

initial velocity of the baseball, u = 27.1 m/s

applied force in opposite direction, F = -5000 N

final velocity in opposite direction, v = -37 m/s

Note: The applied force and final velocity are negative because they act in opposite direction to the initial velocity.

impulse received by the body = change in momentum of the  body

Ft = Δmv

Ft = mv - mu

Ft = m(v-u)

t = m(v-u) / F

Therefore, the bat and the ball were in contact for 1.8 x 10⁻³ s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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d is the correct answer