75.0kg Man decends from 3.1m above the ground. At 3.1m his feet touch the ground and he absorbs the landing through a constant acceleration of 0.6m to come to a verticle rest. What is the net force on him?
Vi = √(2*a*y); a = -[(Vi)^2]/(-2*y); F = m*a; where y = -0.60m and a = -9.8m/s^2.
The Attempt at a Solution
Vi = √[2*a*(-0.60m)] = -7.79m/s
a = -[(-7.79m/s)^2]/(-2*(-0.60m) = -50.63m/s^2
F = 75.0kg*(-50.63m/s^2) = -3800N.
I get 3800N from the ground to him, but the book has 4530N. Their answer works if we add on the absolute value of his mass multiplied by gravity, but why would they do that?