What Is the Average Force Exerted on His Torso by His Legs During Deceleration?

[alexoftennis]

Homework Statement

A person jumps from the roof of a house 4.4 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. The mass of his torso (excluding legs) is 40 kg.

(a) Find his velocity just before his feet strike the ground. ( I found as 9.2865)

(b) Find the average force exerted on his torso by his legs during deceleration. (I found as 2464 BUT THIS IS WRONG. I NEED YOUR HELP TO GET THIS. Thanks!)

Homework Equations

The Attempt at a Solution

(v final) ^2- (v initial)^2 = 2as
a = 9.8 m/s
s=4.4
v initial = 0
solve for v final
v final = 9.2865 m/s

same equation is used in reverse
v final = 0
v initial = 9.2865 m/s
s = 0.70 m this time
solve for a
a = 61.60 m/s

F = ma m is 40kg
F = 40*61.60 = 2464 Newtons.

 

[Fightfish]

2464 N is the net upward force exerted on his torso, not the force exerted on his torso by his legs - his torso still experiences a downward gravitational force.