1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dynamics and Free Fall

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    [tex] \vec{F} = m \vec{a} [/tex]



    3. 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.
     
  2. jcsd
  3. Sep 16, 2007 #2
    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?
     
  4. Sep 16, 2007 #3
    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?
     
  5. Sep 16, 2007 #4
    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.
     
  6. Sep 16, 2007 #5
    So,

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

    Final Answer.

    How's that look?
     
  7. Sep 16, 2007 #6
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Dynamics and Free Fall
  1. Free Fall (Replies: 1)

  2. Free fall (Replies: 5)

  3. Free Fall (Replies: 3)

Loading...