A 0.150-kg baseball traveling in a horizontal direction with a speed of 21.0 m/s hits a bat and is popped straight up with a speed of 15.0 m/s. If the bat was in contact with the ball for 51.0 ms, what was the average force of the bat on the ball

17.65N

Explanation:

We know that the relationship for the impulse and momentum is given as

Ft=mv

given data

mass m=0.15kg

t= 51ms= 26.5/1000= 0.051seconds

velocity v1= 21m/s

Velocity v2=15m/s

substituting in the expression we have

Ft=m(v1-v2)

make F subject of formula we have

F=m(v1-v2) /t

F=0.15(21-15)/0.051

F=0.15*6/0.051

F=0.9/0.051

F=17.65N