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A baseball is moving at a speed of 40.0 m/s toward a baseball player, who swings his bat at it. The ball stays in contact with the bat for 5.00×10−4 seconds, then moves in essentially the opposite direction at a speed of 45.0 m/s. What is the magnitude of the ball's average acceleration over the time of contact? (These figures are good estimates for a professional baseball pitcher and batter.) Find answer in m/s2

 Jul 4, 2016

Best Answer 

 #1
avatar+33653 
+5

acceleration = change in velocity/time taken

 

acceleration = (40 - -45)/0.0005 m/s^2

 

                    = 85/0.0005 m/s^2

 

                    = 170000 m/s^2

 Jul 4, 2016
 #1
avatar+33653 
+5
Best Answer

acceleration = change in velocity/time taken

 

acceleration = (40 - -45)/0.0005 m/s^2

 

                    = 85/0.0005 m/s^2

 

                    = 170000 m/s^2

Alan Jul 4, 2016
 #2
avatar+33653 
+5

I should have had the 45 and 40 the other way round.  I.e the numerator should be 45 - -40.  

Alan  Jul 4, 2016

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