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.