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Homework Help: Grade 11 U Free fall question

  1. Feb 19, 2009 #1
    Hey guys

    need help beginning to solve the following:

    A skydiver accidentally drops a camera out of the plane. The diver notices the mistake 3.0s later and dives out of the plane with a downward velocity of 10.0 m/s. The camera experiences free fall(9.8 m/s^2), but the sky diver accelerates downwards at 8.0m/s^2.

    How far does the camera fall before the sky diver is able to catch it?

    I found how far the camera falls in 3.0s= 44.1 m, and Vf of the camera is 29.4 m/s [down] since Vi is 0. I Don't know really where to go from here, any help will be appreciated!!

    i think the teacher mentioned it takes 39 seconds, but how would i find that
     
  2. jcsd
  3. Feb 19, 2009 #2

    djeitnstine

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    What is true about the displacement when the diver meets the camera?
     
  4. Feb 19, 2009 #3
    displacement would be the same as the camera?
     
  5. Feb 19, 2009 #4

    djeitnstine

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    precisely
     
  6. Feb 19, 2009 #5
    so i need to do d=motion equation of camera=motion equation of diver?
     
  7. Feb 19, 2009 #6

    djeitnstine

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    You've got it.
     
  8. Feb 19, 2009 #7

    djeitnstine

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    If you've done it correctly you should get a quadratic
     
  9. Feb 19, 2009 #8
    ok thank you

    but how do i deal with the 10m/s and the 8.0 m/s^2 when deriving the divers motion equation? they can't simply be added together can they?
     
  10. Feb 19, 2009 #9

    djeitnstine

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    I am assuming that you are using [tex]y=v_{i}t+ \frac{1}{2}at^{2}[/tex] where y is the vertical displacement
     
  11. Feb 19, 2009 #10
    yes.
     
  12. Feb 19, 2009 #11

    djeitnstine

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    Okay so setting 2 equations equal gives [tex]v_{i1}t+ \frac{1}{2}a_{1}t^{2}=v_{i2}t+\frac{1}{2}a_{2}t^{2}[/tex] which gives the quadratic [tex](v_{i1}-v_{i2})t+(a_{1}-a_{2}t^{2})t^{2}=0[/tex]

    So no magic happens =)

    Its up to you to decide which root is relevant

    edit I left out my 1/2
     
  13. Feb 19, 2009 #12
    thank you very much kind sir :)
     
  14. Feb 19, 2009 #13

    djeitnstine

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    You are welcome
     
  15. Feb 19, 2009 #14
    ok so what can i derive from that without t

    edit where did you leave out the 1/2
     
  16. Feb 19, 2009 #15

    djeitnstine

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    1) I placed the 1/2's that you see in the equation.

    2) no t's should disappear since your velocities and accelerations are different
     
  17. Feb 19, 2009 #16
    A simple approach when comparing different velocities is using a table. Of course, I highly recommend working out the mathematics, but a table should also give you the same answer. Example: Car 1: 1s=30m, 2s=50m...Car 2: 1s=45m, 2s=49m....See where they interesect, which is like wording out a quadratic, so-to-speak.

    Regards,

    Fragment
     
  18. Feb 19, 2009 #17
    hmmm
     
  19. Feb 19, 2009 #18
    The solution seems pretty clear to me from this point, djeit pretty much pointed it out. Either graph your quadratics, solve them algebraically, or use the table method.:smile:

    Fragment
     
  20. Feb 19, 2009 #19
    ok, I found the time, 4.5s, using the quadratic formula. it seems wrong, and once i have this time, how do i get displacement?

    which variables do i plug into the displacement formula? my starting values?
     
  21. Feb 19, 2009 #20

    djeitnstine

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    rework your quadratic...i got 10.7777s
     
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