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Kinematics I think it's right, but the computer says NO

  1. Sep 14, 2008 #1
    Kinematics!! I think it's right, but the computer says NO

    An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle theta?

    (the angle is between the ground and the line of sight from the plane to the target. So it is the angle between the adjacent side (the ground) and the hypotenuse of a right triangle with the opposite side being 2.4km)

    There was a diagram with this, but it's pretty self explanatory.

    Vxo = 240(cos(30)) = 207.8 m/s
    Vyo = 240(sin(30)) = 120 m/s

    Vf^2 - Vi^2 = 2a (change in X)
    Vy^2 - 120^2 = 2(9.8)(2400)
    Vy^2 = 247.87 m/s

    (wrong!!)

    and theta:
    the inverse tangent of (247.87/207.8)
    =50.01 degrees

    Also wrong.
    I have no idea where I could have gone wrong. I thought I had this one!

    Please help - I used 4 out of 5 of my attempts and it's due tomorrow!
     
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  3. Sep 15, 2008 #2

    tiny-tim

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    Hi lmbiango! :smile:

    ooh … at last … someone who's tried and who shows us what they've done! Thankyou, lmbiango! :biggrin:

    erm … you're going to kick yourself …

    you've been calculating the angle at which the flare hits the ground, tan-1(Vyf/Vxf) …

    but the question wants the angle from the plane to that point :redface:

    Try again, and show us before you commit yourself! :smile:
     
  4. Sep 15, 2008 #3

    gabbagabbahey

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    Re: Kinematics!! I think it's right, but the computer says NO

    If the flare is falling, shouldn't [tex] v_{yfinal}=-247.87m/s[/tex] ?:wink:

    And if theta is the angle at which the flare hits the ground, then i get:
    [tex] \theta = 50.02^{\circ}[/tex]

    But, if it is the angle between the plane and the line of sight to the target, then I get:
    [tex] \theta =30^{\circ} + 50.02^{\circ}=80.02^{\circ}[/tex]
     
  5. Sep 15, 2008 #4

    tiny-tim

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    Hi gabbagabbahey! :smile:

    No, you're missing the point …

    The flare will fall in a parabola, with a slope that gets steeper.

    The Vy/Vx will be the tan of the slope.

    The line from the plane will be a chord of this parabola.

    You cannot get the slope of that line from the velocities … you must find t first, and then multiply by Vx to get x. :smile:
     
  6. Sep 15, 2008 #5

    LowlyPion

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  7. Sep 15, 2008 #6
    Re: Kinematics!! I think it's right, but the computer says NO

    There is no parabola, the plane just shoots the flare down (as it is landing) and the flare has a curved path from the plane to the target, but never goes up. They want the angle between the target and the ground.
     
  8. Sep 15, 2008 #7
    Re: Kinematics!! I think it's right, but the computer says NO

    I agree, that is another way I tried to do it and got the same answer, but that came up wrong too.
     
  9. Sep 15, 2008 #8

    LowlyPion

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    Re: Kinematics!! I think it's right, but the computer says NO

    They want the angle between the plane at the point of dropping and the ground. That θ is merely line of sight, determined by the tan-1 of height divided by distance.

    Or is there some other angle that θ refers to now?
     
  10. Sep 15, 2008 #9
    Re: Kinematics!! I think it's right, but the computer says NO

    I don't understand... I don't know how to describe this angle to make sure we are talking about the same one- It is the same angle I described to you earlier. Ok there is a triangle. The hypotenuse is your line of sight from the plane to the ground. the top of the triangle is the plane. the right angle point is directly under the plane and is the ground. The other angle (the target) is theta. The opposite side is given = 2400 meters. And the adjacent side, or the distance I calculated using the quadratic equation.

    So by the methods you gave me are you saying that the path of the flare is irrelevant? I have the answer, (still not sure if it's right) but I don't really know how I got it because I don't see how the path of the flare correlates to this right triangle we are talking about. As I said before, the path of the flare is a curve above the hypotenuse with the same endpoints.

    I just want to understand how this works enough to get the answer on my own next time, and to make sure it's right before I get this HW problem wrong.
     
  11. Sep 15, 2008 #10

    tiny-tim

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    Hi lmbiango! :smile:

    (what quadratic equation? can you show us?)

    LowlyPion and you are describing the same angle!

    Its tangent, as both you and LowlyPion say, is 2400 divided by the horizontal distance, xf.

    But have you actually found xf? … if you have, you haven't yet told us.
     
  12. Sep 15, 2008 #11

    LowlyPion

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    Re: Kinematics!! I think it's right, but the computer says NO

    That should have been found by the 240*Cos30 times Time to hit ground.

    I didn't look to see if it was still there but at one point he knew this.

    (See also: https://www.physicsforums.com/showthread.php?t=255677 )
     
  13. Sep 15, 2008 #12

    tiny-tim

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  14. Sep 15, 2008 #13
    Re: Kinematics!! I think it's right, but the computer says NO

    ok so I solved for time in this equation:
    x=Vi(t)+(1/2)(a)(t)^2

    but I am confused on where it came from. The equation my prof. gave me is:
    Xf=Xi+Vi(t)+(1/2)(a)(t)^2
    so wouldn't Xi be 2400m and Xf would be the ground or 0?
    Or if Xi is zero, Xf would be -2400. Either way, Xf would be negative, I would think.

    so by the equation you gave me earlier we have:
    2400 = (240)(1/2)(t) + (1/2)(9.8)(t^2)

    I don't know why you say to multiply the first "t" term by (1/2)

    If it was me coming up with this without your help I would use the equation my professor gave us and I would have:

    Xf=Xi+Vi(t)+(1/2)(a)(t)^2

    -2400 = 240t + (1/2)(9.8)(t^2)

    So, as we know- the answer when I solve it my way is wrong.
    I solved it with your equation: 2400 = (240)(1/2)(t) + (1/2)(9.8)(t^2)
    120 +/- [(square root of) (120^2 - 4 x 4.9 x -2400)] / 2(4.9)

    The only positive value comes out to be 13.0481 seconds

    Now, I multiply that by the x component of my velocity because (I just understood this)
    V= D/T
    So V x T = D and we are talking about x so it is horizontal distance.

    So X= 2711.395 m

    Great, and Inverse tangent of (2400/2711.395) = 41.514 degrees.

    I submitted it and it's right. But I want to understand it.
    Why did you multiply the T by (1/2) and why is your Xf not Yf and why is it not negative?
     
  15. Sep 15, 2008 #14
    Re: Kinematics!! I think it's right, but the computer says NO

    I'm sorry. Forget the other thread. I have never used this site before and I couldn't find where my question from three days ago went, so I started a new one last night. I thought it got rid of my old one, I guess not, but anyway read what I wrote above and tell me what you think.
     
  16. Sep 15, 2008 #15

    LowlyPion

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    Re: Kinematics!! I think it's right, but the computer says NO

    The 1/2 is Sine30. That's your component of y velocity that you have to figure in.

    As to my signs, I didn't think about it. You know the distance it will travel is given by the initial velocity times time plus the distance traveled with the acceleration component. Personally I get confused trying to plug things in some general equation that takes into account everything, because then you have to remember what the direction and assumptions are for that general equation. I just find it simpler to construct the equations from what I need to properly describe it.

    I don't necessarily recommend this approach. It's just the way I roll.
     
  17. Sep 15, 2008 #16

    LowlyPion

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    Re: Kinematics!! I think it's right, but the computer says NO

    Well ... I tried to warn you earlier.

    https://www.physicsforums.com/showpost.php?p=1874391&postcount=5

    A moderator might well have joined the threads and may still to avoid same.
     
  18. Sep 15, 2008 #17
    Re: Kinematics!! I think it's right, but the computer says NO


    OHH! ok I get it. I don't know if I could ever do it by myself, but I get it for now. Thanks for all the help!

    P.S I am a girl
     
  19. Sep 15, 2008 #18

    LowlyPion

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    Re: Kinematics!! I think it's right, but the computer says NO

    Ooops.
     
  20. Sep 15, 2008 #19

    tiny-tim

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    Hi LowlyPion! :smile:

    Yes you did … I got confused by the appearance here of gabbagabbahey, and didn't realise both threads had the same OP! :redface:
    p.s. i am a goldfish :wink:
     
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