1. The problem statement, all variables and given/known data A 231.0 g ball is dropped from a height of 2.60 m, bounces on a hard floor, and rebounds to a height of 1.37 m. The impulse received from the floor is shown below. What maximum force does the floor exert on the ball if it is exerted for 2.00 ms? 2. Relevant equations Jx = [tex]\Delta[/tex]p Fmax*[tex]\Delta[/tex]t = Jx 3. The attempt at a solution so i though that Jx = [tex]\Delta[/tex]p = mvf - mvi but for this question you have to add the momentums.. is that because at first the momentum is pointing down and then after the collision the momentum is pointing up so they have opposite signs?? thanks for the help!