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A plane flying at 220m s-1 at a height of 300m

  1. Oct 7, 2012 #1
    This is a question about projectiles

    A plane flying at a constant speed of 220m s-1 at a height of 300m. The pilot drops a food package to hit a certain target.

    a)How long (in time) before passing over the target should the pilot release the package?
    b)How long in metres would that be?



    S = ut + at^2/2

    This is not my attempt at the question but someone else's but I don't trust his answer as I'm unsure about it although it looks right.


    S = 300
    a = 9.8
    u = 220

    300 = 220t + 9.8t^2/2
    4.9t^2 + 220t - 300 = 0

    After solving using the q. Formula he got t as 1.45 s

    b) S = ut
    S = 220x1.45
    S = 319m




    Is his answer to this question right? It sure looks like it to me but I don't understand why , the package's initial vertical velocity is 0 so its horizontal velocity 220 m s-1? Or is that the plane's?
    1.45s before they pass over the target they should release the package? What would happen if they released it just as they were over the target?

    Thanks
     
  2. jcsd
  3. Oct 7, 2012 #2
    a)
    S = ut + (at^2)/2

    S = 300
    a = 9.8
    u = 0 <----because VERTICAL initial velocity is 0 (unless your plane is flying towards the ground...)

    300=0.5(9.8)t^2
    t=7.82s (3 s.f)

    b)s=ut
    s=220x7.82
    s=1720.4m
     
  4. Oct 7, 2012 #3
    I thought this was this answer too , are you sure that it's right can anyone else verify?
     
  5. Oct 7, 2012 #4
    I'm 90.5% sure about it haha :p
     
  6. Oct 7, 2012 #5
    A detailed explanation of why should help me understand , I don't know who is right about it now.
     
  7. Oct 7, 2012 #6
    What detailed explanation do you need? The vertical velocity is indeed zero initially, which gives a little less than 8 s of free fall.
     
  8. Oct 7, 2012 #7
    When dealing with 2D vectors like this, you can look at each direction seperately.
    The vertical component is not affected by the horizontal component which is directly perpendicular to it.
     
  9. Oct 7, 2012 #8
    Yes I know that it is 0 but is that what the question is asking ? To me it says how long before they reach the target should they release the package , doesn't it mean that it's not falling straight down as if from rest (where both initial horizontal and vertical are 0) but 'diagonally'?
     
  10. Oct 7, 2012 #9
    ya its diagonally. but the 220ms-1 only affect how fast it moves horizontally.
    The diagonal component can be seen seperated as vertical and horizontal
     
  11. Oct 7, 2012 #10
    It may be moving diagonally, but the gravity affects only the vertical component of the velocity, so you can solve for it independently.
     
  12. Oct 7, 2012 #11
    Sorry but I still don't understand , I also got 7.8 seconds when I worked it out but wasn't the time taken for the package to hit the ground ? Surely that's different from the time they have to release it before they fly over the target?
     
  13. Oct 7, 2012 #12
    Why is that different and especially why is that SURELY different?
     
  14. Oct 7, 2012 #13
    Hard to phrase now that I think it , the time it takes to hit the ground is 7.8 seconds (apparently) if they released it anywhere because the initial vertical velocity would still be 0 . The time before they should release it before they pass over the target since they are travelling at 220 m s-1 , do you see what I mean by that

    Is the packages horizontal velocity 220m s-1?
     
  15. Oct 7, 2012 #14
    Ya, you think we should find the time taken to reach the horizontal distance right? :)
    You have to imagine here. Since in reality the object is dropping diagonally, it is falling vertically and horizontally simultaneously. The time taken to reach the vertical distance will be the same to reach the horizontal distance.
     
  16. Oct 7, 2012 #15
    It takes 7.8 s to hit the ground no matter how fast the plane is flying. It depends only on the altitude.

    However, that need not be the time before the fly-over. The drop time should be computed from the requirement that the distance to the target at the drop time equal the horizontal velocity times 7.8.
     
  17. Oct 7, 2012 #16
    Nubcake, draw it out and u'll find it clearer :)
     
  18. Oct 7, 2012 #17
    I tried drawing it out but didn't make much sense , so let me get this straight , the correct answer is that they should release it 7.8 seconds before they fly over the target since it has a horizontal velocity of 220m s-1? Then it will land on the target otherwise if they dropped it whilst over the targe it wouldn't land on it but away from it?
     
  19. Oct 7, 2012 #18
    ya you can say that :). just that from this case we deduce the time taken first, then deduce the appropriate horizontal distance
     
  20. Oct 7, 2012 #19
    Alright then I'll accept your answer for now until I check with my teacher:smile:
     
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