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Magnitude of Acceloration problem

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    a person jumps from a window 46m high. and is caught in a figherfighters net which streaches .6m. to what tenth of a m/s^2 is the persons magnitude of acceloration in the net


    2. Relevant equations
    X= - 46
    a = -9.8
    v= ?



    3. The attempt at a solution
    would the soultion be -9.8? im really not sure what formula to use or how to solve this. PLEASE HELP!
     
  2. jcsd
  3. Nov 9, 2008 #2
    Is this all of the given information? Something seems to be missing.
     
  4. Nov 9, 2008 #3
    yeah, thats all. the hit was that theres 2 parts to the problem, falling and hitting the net, idk if that helps at all...
     
  5. Nov 9, 2008 #4
    IDK why she's in the 10th floor jumping into a net but...

    Can you please write every single word, as written, with the question clearly indicated? I have no idea what needs to be solved for.
     
  6. Nov 9, 2008 #5
    A person jumps from a window 46 meters high and is caught in a firefighter's net which stretches 0.6 meters. to the nearest tenth of a M/S^2, what is the magnitude of the persons acceloration in the net.

    that is copy/pasted from my hw.
     
  7. Nov 9, 2008 #6
    Okay, well, here's a stab at it:

    [tex]v_f^2=v_i^2+2ad \rightarrow v_f=\sqrt{92g}[/tex]

    From this, we'll now solve for the second portion of motion:

    [tex]v_f^2=v_i^2+2ad \rightarrow -\sqrt{92g}=\sqrt{1.2a}[/tex]...
     
  8. Nov 9, 2008 #7
    could u possibly show me the variables u used to substitute into the formula?
    im kinda having a hard time with this, and im not exacly following what your doing...
     
  9. Nov 9, 2008 #8
    The square of your final velocity during free fall is equal to your initial velocity-squared (0m/s) plus two times the quantity of your acceleration due to gravity, g=9.81m/s/s, and your displacement, 46m. This allows you to solve for your final velocity to be, 30m/s. Then, you do the same thing for the second part. Your final velocity-squared (0m/s) is the square of your initial velocity, 902.52 plus twice your acceleration due to the net, a, and your displacement, 0.6m. Rearranging, you'll solve for a, giving you a=752m/s/s, I think. I'm rushed, calculations might be wrong.
     
  10. Nov 9, 2008 #9
    Thank you so much!!
     
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