SOLVED 1. The problem statement, all variables and given/known data Let me start by saying my ONLY question is. Why can I not use.. vf=(2aΔs).5 Why do I have to use vf=(2mgH).5 to get the vf? Why have things changed now that I am dealing with momentum problems, isn't vf just vf? With the mgH formula my velocity is much lower. This doesn't make sense and its bothering me that I am missing something so fundamental. But I'll post the problem anyway. A 160 g ball is dropped from a height of 2.8 m, bounces on a hard floor, and rebounds to a height of 2.1 m. Figure P9.28 shows the impulse received from the floor. What maximum force does the floor exert on the ball? http://img441.imageshack.us/img441/8416/p928.gif [Broken] 2. Relevant equations vf=(2aΔs).5 (I thought but I guess not) vf=(2mgH).5 Jx=Δpx=FmaxΔt 3. The attempt at a solution I am solving it using the method written in the link below. Which brought me to the question listed above. http://answers.yahoo.com/question/index?qid=20081026102629AAdYIDa """"" You start by defining what "impulse" is. In math talk, it's dP = F dt; where dP = p1 - p0 = m(v1 - v0), the change in momentum, F is the average force over the time interval dt = t1 - t0 in which the momentum changed. F = ? is the force you are looking for. Solve for F = m(v1 - v0)/dt; where m = .15 kg, dt = .005 sec, you need to find v0, the velocity at impact and v1 the velocity at rebound. To find v0, use the conservation of energy so that v0^2 = 2mgH; where H = 1.7 m. Then v0 = sqrt(2mgH) = sqrt(2*.15*9.81*1.7) = 2.236761051 mps. Here the potential energy mgH is converted to kinetic energy. To find v1, use the same conservation of energy so that v1 = sqrt(2*.15*9.81*1) = 1.715517415 mps, where the height h = 1.0 m on rebound. Here the kinetic energy of the rebound is converted to potential energy PE = mgh at height h. Recognize that the directions of the two velocities are directly opposite; so they are additive, in other words v1 - (-v0) = v1 + v0 = 3.952278466 Solve for F = dP/dt = m(v1 + v0)/dt = .15*3.95/.005 = 118.5 kg.m/sec^2. """""