A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51m/s to the north. What was the average acceleration of the ball during the 1.0ms when it was in contact with the bat?
Answers & Comments
uneq95
Let us assume the south direction as positive and the North direction as negative.
Acceleration at the point of contact = (43-(-51))/0.001 s = 94000m/s²
I hope you understand it well.
Please mark it as brainliest if you find it helpful. Gud luck!
Answers & Comments
Acceleration at the point of contact
= (43-(-51))/0.001 s
= 94000m/s²
I hope you understand it well.
Please mark it as brainliest if you find it helpful.
Gud luck!