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Equal Weights problem.

  1. Oct 3, 2009 #1
    problem is attached

    so far i've found all the variables i could, and i've noticed that the vertical acceleration is 1.5 times larger in the vertical direction when the propeller is on than when the aircraft is flying. there's just so many different variables and situations and everything that i don't know where to get started.
     

    Attached Files:

    Last edited: Oct 3, 2009
  2. jcsd
  3. Oct 3, 2009 #2
    The attachment is still pending approval.
     
  4. Oct 3, 2009 #3
    okay, fixed that till it get approved.

    image is now in thumbnail
     
  5. Oct 3, 2009 #4
    Wow, that is one of the most horribly worded problems ever.

    Also I can't see a way of doing that. In that particular system the spacecraft would remain under the aircraft from what I see.
     
  6. Oct 3, 2009 #5
    yeah, it would, but then once the propellor kicks in it starts accelerating to get ahead of the aircraft.

    and i agree, my prof really could've made this a lot less confusing by writing it out better =\
     
  7. Oct 3, 2009 #6
    Oh wait, I just re-read it. So all you need to do is draw a triangle. You know all the angles, and one of the sides. That's enough to solve for the rest.

    Or try using components. Get the X and Y components of the thrust, then see how long it would take to take to move 1km forward, then use that time to work out the amount it has to fall.
     
  8. Oct 3, 2009 #7
    Okay, since the speed of the spacecraft is 0m/s relative to the aircraft, i can find the final velocity(horizontal) of the spacecraft relative to the aircraft. this is ~225m/s, which it takes 8.86s to get 1000m in front of the aircraft... so then what?
     
  9. Oct 3, 2009 #8
    Now you solve for the Y part of it. Get the Y component of it's velocity at that point. Then use the time to backtrack and see just how far it has to climb to get to the speed at 1km in front of the aircraft.

    Once you have the height relative to the airplane (it should be negative) then all you have to do is work a simple equation to see how long it would take to all that distance on it's own.
     
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