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Projectile range

  1. Dec 31, 2013 #1
    1. The problem statement, all variables and given/known data

    After a short engine firiing, an atmosphere-probing rocket reaches 4.6km/s. If the rocket must land within 50km of its launc site, what's the max allowable deviation from a vertical trajectory.

    2. Relevant equations
    none


    3. The attempt at a solution

    Given by the book:

    x = (vi^2/g)sin 2Θ

    50,000m = (vi^2/-9.8)sin 2Θ

    sin2Θ = (gx/vi^2) = 0.0232

    Again this is really frustrating, I take g = -9.8 and x = 50,000m, vi = 4600m/s
    but I get -0.0232.

    What is wrong here?
    If I use a positive g, I get 0.0232 but that would be rubbish.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 31, 2013 #2

    Curious3141

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    Why would it be "rubbish"?

    I suggest you check the derivation of the range formula. That should tell you why ##g## should be taken as a positive number (just the magnitude) here.
     
  4. Dec 31, 2013 #3
    I understand how to derive the equation for the range formula. What I don't understand is that the g in the equation was not specified as a magnitude so why should I be using the magnitude and not the vector g?
     
  5. Jan 1, 2014 #4

    Curious3141

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    So, in the derivation, what is the exact expression for the vertical position at time ##t##?
     
  6. Jan 2, 2014 #5

    Capture.JPG

    I have understood why g = 9.8 instead of -9.8ms^-2
     
  7. Jan 2, 2014 #6

    Curious3141

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    Good. It is because the equation for the vertical displacement explicitly includes a negative sign before ##g##, i.e. ##\displaystyle y = v\sin\theta{t} -\frac{1}{2}gt^2##. Hence ##g## here refers to the magnitude only.
     
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