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Motion of a projectile thrown from a plane

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    A helicopter is flying horizontally at 8.5 m/s and an altitude of 18 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 10 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?


    2. Relevant equations
    Not quite sure.


    3. The attempt at a solution
    None. I'm thinking I have to find time first but I'm not sure how to do that either.
     
  2. jcsd
  3. Sep 20, 2007 #2
    To find time, it would be T=sqrt(2h/g) right. I did that then did d=vt for both the helicopter and the package thrown and added them together but still didn't get the right answer.

    Any suggestions?
     
  4. Sep 20, 2007 #3

    Doc Al

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    Why did you add them? The package is ejected backwards. (The easy way is to find and use the horizontal speed of the package with respect to the ground.)
     
  5. Sep 20, 2007 #4

    mgb_phys

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    Sounds right. I get t=1.9sec to reach ground
    Package is moving backward at (10-8.5) m/s the helicopter is moving forward at 8.5 m/s
    so total distance is (8.5+10-8.5)*1.9 = 19m

    You can simply pick the helicopter as the frame of reference, then it is just thrown out at 10m/s for 1.9s
     
  6. Sep 20, 2007 #5
    I added the two distances because I thought that would give the total distance.
     
  7. Sep 20, 2007 #6
    Thanks. I see what I did wrong and why I wasn't getting the right answer.
     
  8. Sep 20, 2007 #7

    Doc Al

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    What if the package were ejected at a speed of 8.5 m/s backwards with respect to the plane? What would the net horizontal distance traveled be then?
     
  9. Sep 20, 2007 #8

    mgb_phys

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    As I said, it's a bit of tricky question ( or a good intro to relativity)
    Because it is thrown backward tou have to subtract the forward speed of the helicopter but if you are measuring from the groudn you then add the speed of the helicopter back in - they cancel. Consider viewing it from the helicopter (or imagine the helicopter is hovering) then it is just thrown backward with 10m/s.
     
  10. Sep 20, 2007 #9

    Doc Al

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    You are miscalculating the speed. The package is moving at -10 m/s relative to the plane; the plane is moving at +8.5 m/s relative to the ground. The speed of the package relative to the ground is -10 +8.5 = -1.5 m/s backwards.
     
  11. Sep 20, 2007 #10

    mgb_phys

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    Yes but the helicopter is still moving 8.5m/s forward (relative to ground) after the drop so the separation on the ground is increasing at 10m/s
     
  12. Sep 20, 2007 #11

    Doc Al

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    D'oh! I misread the question. :redface: Sorry about that! You are absolutely correct.
     
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