(adsbygoogle = window.adsbygoogle || []).push({}); 1. A ball is dropeed from a height of 2.5m and rebounds to a height of 2.1m. If the ball is in contact with the floor for .70ms determine 1: direction and 2: the magnitude of the balls average acceleration due to the floor

I know the equation for average acceleration = change in velocity/ time to make change

I found initial velocity by: v(initial)= sqrt(2*9.8*2.5) = 7m/s

I found final velocity by: v(final)= sqrt(2*9.8*2.1) = 6.42m/s

When i plug those into the avg acceleration equation i have: (6.42m/s - 7 m/s)/.0007s

to get: -828.57m/s So the direction would be down?

Am i doing my math right I am not sure where the negatives go for gravity either. Thanks for your help.

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# Homework Help: Direction and magnitude of the average acceleration

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