perhaps the question is:
at what speed does she leave the ground? Hint: The force of the floor is not the only force acting on the woman.
from the graph at t =0, the floor exerts an upward force on the lady = to her weight = 600 N = mg
=> m = 600/g = 600 / 9.81 = 61.16 kg
F̂ₐᵥₑΔt = Î = Impulse = change in momentum = Δ(mv̂)
here Fₐᵥₑ(net)Δt = area of the graph with the x-axis – impulse due to her weight(downwards)
= 600(0.2) + ½(0.275)1800 + (0.275)600 + ½0.025(600) – 600(0.5)
= 240 N
therefore 240 = mΔv = 61.16 [v – 0]
=> v = 240/61.16 = 3.896 m/s
hope this helps
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