1. The problem statement, all variables and given/known data When a person parachutes, the impact velocity is equal to that attained in free fall from a height of 4.5 m. After contacting the ground, the jumper’s momentum is quickly brought to zero by the earth. Assuming that the person has a mass of 80 kg, what is the average force on the jumper’s feet if he lands stiff-legged and the impulse lasts only 0.019 s? 3. The attempt at a solution v22 = v12 + 2aΔd v22 = (2)(1.8)(4.5) v2 = 9.4 m/s [down] p = mv p = (90)(9.4) p = 752 N F = Δp/Δt F = 750/0.019 F = 3.9 × 104 N [up] However, the correct answer is 4.03 x 104 N [up]. My teacher hinted towards the normal force to the person's weight is added to the answer. This confuses me as I thought that the impulse takes that into consideration already. Someone please help clarify this for me. Thanks in advance.