# Net Force Question

## Homework Statement

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?

## Homework Equations

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?

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## Homework Statement

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?

## Homework Equations

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?