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fast ball with a velocity of 43 m/s south. The batter hits the ball and gives it a velocity of 51 m/s north. What was the acceleration of th
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fast ball with a velocity of 43 m/s south. The batter hits the ball and gives it a velocity of 51 m/s north. What was the acceleration of the ball if it was in contact with the bat for 1 s? What direction did it go?
PLEASE HELP I NEED THIS DONE FOR MY NEXT CLASS
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Physics
3 years
2021-07-16T00:27:48+00:00
2021-07-16T00:27:48+00:00 1 Answers
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Answers ( )
Acceleration = (change in velocity) / (time for the change)
change in velocity = (velocity after) – (velocity before)
change in velocity = (51 m/s north) – (43 m/s south)
that’s the same thing as (51 m/s north) + (43 m/s north)
change in velocity = 94 m/s north
Acceleration = (94 m/s north) / (1 second)
Acceleration = 94 m/s² north
The direction of the acceleration is North.