A person jumps from the roof...Calculating Leg Force during Deceleration

[esong]

A person jumps from the roof...

Homework Statement

A person jumps from the roof of a house 3.9 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.73 m.

- PART A If the mass of his torso (excluding his legs) is 40 kg, find his velocity just before his feet strike the ground.

- PART B: If the mass of his torso (excluding his legs) is 40 kg, find the magnitude of the average force exerted on his torso by his legs during deceleration.

Homework Equations

The Attempt at a Solution

I've done PART A, and got 8.7 m/s. (or \sqrt{2*9.8*3.9}))
I tried to get the acceleration during the deceleration by using the equation
(vf)^2 = (v0)^2-2ad
2a(.73m) = (8.7m/s)^2
a = 52.35
And multiplied the acceleration by the mass (40 kg) to get 2094 N,
but the program says I'm wrong! What am I doing wrong here?

 

[Cynapse]

As far as I can see you are putting the numbers in wrong...

I get
(vf)^2 = (v0)^2-2ad
0 = -(8.7m/s)^2 - 2a(.73m) <= REMEMBER that vf is 0 and vi is 8.7
2a(.73m) = -(8.7m/s)^2
a = -(8.7^2)/(2*0.73)
a = -51.8424658
40a = -2073.69863N

 

[PhanthomJay]

I don't know about the numbers, but you are neglecting the weight of the person during impact. 2100+/- Newtons is the net force acting during the deceleration. That is not the the 'leg' force.

 

[esong]

Sorry, but that's still the wrong answer.
Besides, the answer is a positive number.

 

[PhanthomJay]

Sorry, but that's still the wrong answer.
Besides, the answer is a positive number.

Your acceleration value is OK, but you won't get the right answer for the leg force until you draw a free body diagram of the person's torso and note the forces acting on it. One of those forces is the force of his legs on the torso. What's the other?? Then, use Newton's 2nd Law to solve for the leg force. Mind your force and acceleration directions.