What is the Force Exerted on a Parachutist During Free Fall Deceleration?

[Jacobpm64]

Homework Statement

A parachutist of mass 80 kg approaches the ground at 5 m/s. Suppose that when he hits the ground, he decelerates at a constant rate (while his legs buckle under him) over a distance of 1 m. What is the force the ground exerts on his feet during the deceleration?

Homework Equations

[tex] \vec{F} = m \vec{a} [/tex]

The Attempt at a Solution

I think a lot of this information is extraneous. Isn't the force that the ground exerts on the parachutist going to be opposite his weight?

So, [tex] \vec{F} = mg = (80 * 9.8) = 784 [/tex] N

I'm not sure about this because it seems odd that all of that extra information would be given if it is not needed.

Thanks in advance.

 

[bob1182006]

I think you're supposed to find the acceleration (decelleration in this case) before you find the force the ground exerts on him.

You're given a velocity, a distance, what equation can you use to determine the decelleration?

 

[Jacobpm64]

So, let's see what I can do.

[tex] x_{0} = 1 m[/tex]
[tex] v_{0} = 5 m/s[/tex]
[tex] x = 0 m[/tex]
[tex] v = 0 m/s[/tex]

[tex] a(x-x_{0})=\frac{1}{2}(v^2 -v_{0}^2) [/tex]

solving for a,

[tex] a = \frac{v^2-v_{0}^2}{2(x-x_{0})} [/tex]

Plugging in the numbers,

[tex] a = \frac{(0)^2 - (5)^2}{2(0-1)} [/tex]

[tex] a = 12.5 m/s^2 [/tex]

So, what do I do now, do I still factor in gravity?

 

[bob1182006]

ok so decelleration is 12.5m/s^2 now you can use that a in order to find the force he exerts on the ground, and then the force the ground exerts on him.

 

[Jacobpm64]

So,

[tex] F = ma = (80 * 12.5) = 1000 N[/tex]

Final Answer.

How's that look?

 

[bob1182006]

yes but the acceleration should be -12.5 since it's deceleration,
so the parachuter exerts -1000N on the ground, and the ground exerts the 1000N on the parachuter.