1. The problem statement, all variables and given/known data mass of steel ball = 40g = .04 kg h1 = 2.00 m h2 = 1.60 m A steel ball is dropped from height of 2.00 m onto horizontal steel slab, rebounds to height of 1.60m. Calculate impulse of ball during impact. 2. Relevant equations impulse J = F*dt J = P2 - P1 K0 + U0 + Wother = Kf + Uf 3. The attempt at a solution I tried to get velocities using conservation of energy in two steps; from drop to impact; from impact to rebound. .04kg(9.8m/s2)(2m) = 1/2(.04kg)v2 v = 6.26m/s 1/2(.04kg)v2 = .04kg(9.8m/s2)(1.6m) v = 5.6m/s then I used ΔP = .04kg(5.6m/s - 6.26m/s) = -.0264 Book shows answer to be .47N*s Not sure where I'm botching this up. Any help is appreciated.