What Is the Correct Way to Calculate Average Force in a Jumping Scenario?

[sunnyday]

Homework Statement

A person jumps from the roof of a house 4.2-mhigh. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.66 m .

If the mass of his torso (excluding legs) is 43 kg , find the magnitude of the average force exerted on his torso by his legs during deceleration.

Homework Equations

v_y² = v₀² +2ad
F=ma

The Attempt at a Solution

I already got 9.077 as v₀ (this is correct)
v_y = 0
d = 0.66

0 = 9.077² + 2a(0.66)
-82.391 = 1.32a
a = -62.41F = 43(-62.41)
F = -2683.97 (this answer is wrong and I don't know why)

 

[Jamison Lahman]

Perhaps they want you to find the force of gravity during the deceleration and add that?

 

[sunnyday]

Perhaps they want you to find the force of gravity during the deceleration and add that?

I don't know how to do that.

 

[Jamison Lahman]

I don't know how to do that.

$F=m(a+a_g)$
Also, I believe magnitude implies a positive number.

 

[haruspex]

I don't know how to do that.

What would the force on his torso have been if he had jumped from only .0001m? Or if he'd not jumped at all, but just stood there?

Soapbox: The question is wrong. There is no way to answer it with the information given.
Average force is defined as Δp/Δt, the change in momentum divided by the elapsed time. It cannot in general be calculated by ΔE/Δs, the change in energy divided by the distance the force moves. They will give the same answer if the force is constant, but otherwise they may be different. In the specific case of a person landing from a jump, the force will be small initially and increase as the muscles tauten.

Instead of asking for average force, the question should say "what is the force, assuming it is constant?"

See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ for a longer rant.