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Finding Time, Distance, and Velocity

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 300m. You can ignore air resistance.

    How much time is required for the bomb to reach the earth?
    How far does it travel horizontally while falling?
    Find the horizontal and vertical component of its velocity just before it strikes the earth.



    2. Relevant equations

    Velocity Final = Velocity Initial + Acceleration(Time)
    X Final - X Initial = Time/2 (Velocity Initial + Velocity Final)
    X Final - X Initial = Velocity Initial (Time) + .5 (Acceleration)(Time)^2
    Velocity Final^2 = Velocity Initial ^2 + 2 ( Acceleration)(X Final - X Initial)

    3. The attempt at a solution

    I want to really understand this, so I will be breaking up the work into parts. As usually the result of the previous question help solve the one after it.

    For the first question:
    How much time is required for the bomb to reach the earth?

    The knowns: Final Position @ 300 and Initial @ 0. Acceleration @ -9.8. Velocity Initial @ 60.
    The unknown: Time

    Based on the values provided, the following equation seems suited to the problem.

    X Final - X Initial = Velocity Initial (Time) + .5 (Acceleration)(Time)^2

    0 - 300 = 60(t) + .5 (-9.8)(t^2)
    -4.9t^2 +60t +300 = 0

    -60 +- 97.365 / -9.8

    t = 16.06
    or
    t = -3.81

    This incorrect. How would I have to approach this problem.:bugeye:
     
  2. jcsd
  3. Sep 16, 2007 #2

    Kurdt

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    You have to think about this problem in two dimensions. To find the time you are along the correct lines. You know it will experience a constant acceleration until it hits the ground 300 m below. But the vertical component of velocity for the bomb when it is released is 0 m/s not 60 m/s.

    Also remember your units.
     
  4. Sep 16, 2007 #3
    So the knows would have to be:

    The knowns: Final Position @ 300 and Initial @ 0. Acceleration @ -9.8. Velocity Initial @ 0

    0 - 300 = 0(t) + .5 (-9.8)(t^2)
    -4.9t^2 = -300
    t^2 = 61.22
    t= 7.82

    also, since it's 300 below , would it not be -300?
     
  5. Sep 16, 2007 #4

    Kurdt

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    Yes that looks ok. Now how about part two.
     
  6. Sep 16, 2007 #5

    Kurdt

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    Sorry missed this bit.

    It all depends how you are defining things. All that matters is that you are consistent. You could for instance say the ground is at 0m and the helicopter is at 300m or the other way round. Or the helicopter is at 0m and the ground is -300 m.
     
  7. Sep 16, 2007 #6
    yes, but here I am placing the "origin" at the helicopter, so the ground is -300 downwards. Hence why I am confused why the equation does not look like this instead:

    0 "+" 300 = 0(t) + .5 (-9.8)(t^2)
     
  8. Sep 16, 2007 #7

    Kurdt

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    Well in that case you'd have to define the acceleration downwards as a positive quantity otherwise the equation won't make much sense.
     
  9. Sep 16, 2007 #8
    How far does it travel horizontally while falling?

    The knowns: Final Position @ 300 and Initial @ 0. Acceleration @ -9.8. Velocity Initial @ 0. Time final = 7.82.

    Horizontally means the x axis.

    If the object falls, would it not just fall down straight. (maybe the little push from the copter)

    Either way:

    X Final - X Initial = Time/2 (Velocity Initial + Velocity Final)
    X Final - 0 = 7.82/2 (60+ 0)
    ?
     
    Last edited: Sep 16, 2007
  10. Sep 16, 2007 #9

    Kurdt

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    The initial velocity in the horizontal direction is 60 m/s and so is the final velocity. The reason this doesn't change is because the question asks us to ignore air resistance.
     
  11. Sep 16, 2007 #10
    Well just a bit ago, I realized that horizontal velocity is 60, but why is the final velocity 60 still. I would think that once it bombards into the ground, velocity would be 0.
     
  12. Sep 16, 2007 #11

    Kurdt

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    Yes once it hits the ground it will be zero, but the equations you are using will assume that it undergoes some uniform decelleration if you assume the final velocity is zero which it doesn't. So we say it hits the ground in 7.82s in which time its travelling at a constant horrizontal speed of 60m/s.
     
  13. Sep 16, 2007 #12

    Kurdt

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    Just going back to this for a second. That would be incorrect, since the equations you are using state [itex] x_{final} - x_{initial} = ... [/itex] and thus you'd get -300 + 0 = -300.
     
  14. Sep 16, 2007 #13
    so it would be more like this:

    X Final - 0 = 7.82/2 (60+ 60)
    X Final - 0 = 7.82/2 (120)
    469.2 = x final
     
  15. Sep 16, 2007 #14

    Kurdt

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    That looks correct.
     
  16. Sep 16, 2007 #15
    ok so we know that the horizontal component prior to reaching ground is 60m/s. and that the vertical when prior to reaching ground is 0 right..

    nvm the vertical

    i will need to solve using difference in distance over time.
     
  17. Sep 16, 2007 #16

    Kurdt

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    Just use the first equation you have in the first post to find the vertical component before impact. You know the initial velocity is 0 m/s.
     
  18. Sep 16, 2007 #17
    the first equation I have in the first post was : Velocity Final = Velocity Initial + Acceleration(Time)

    how does this determine max height?

    anyways here it goes

    vf = 0 + -9.8(3.41)

    -33.418.......


    hehehe...I noticed that I mixed it with another problem.
     
    Last edited: Sep 16, 2007
  19. Sep 16, 2007 #18

    Kurdt

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    I thought the last question was concerned with what the horizontal and vertical components of velocity were. I also thought we had determined the time to be 7.82 s.
     
  20. Sep 16, 2007 #19
    yes time is 7.82 for the bomb to reach earth, and we are trying to find the horizontal and vertical components of velocity of bomb just before it hits earth. You helped me determine that the horizontal is 60 m/s.

    We are trying to figure out the vertical.

    and you also pointed out to use the first equation...

    I just did, and I get the result of -76.63

    v=-9.8 (7.82)

    redoing a^2+b^2 =c^2

    a^2 + 60^2 = 76.63^2
    a= 47.68

    it gives me 47.68

    which is wrong. :( only got one more try. :(
     
    Last edited: Sep 16, 2007
  21. Sep 16, 2007 #20

    Kurdt

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    I don'y understand what you've done. The question asks for what the components are. You seem to be trying to work out the final velocity before impact. Are you sure the question is as you have stated it?
     
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