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Why is magnitude g instead of vector g used in range projectile

  1. Dec 31, 2013 #1
    It is not being explained anywhere in my book nor on the internet.
    Can someone explain to me why?
     
  2. jcsd
  3. Dec 31, 2013 #2

    Bandersnatch

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    It's because in projectile motion treatment, the positive value of displacement is assumed to be "up", and accordingly all derivatives have the same assumption embedded in them. V is "up" and -V is "down". a is "up" and -a is "down".
    On the surface of Earth, the magnitude of the acceleration vector is assumed to be g, and it's direction "-", or "down", i.e., towards the negative values.

    You can use proper vector notation and get the same results. It's just that most textbooks tackle projectile motion before properly explaining vectors.
     
  4. Dec 31, 2013 #3
    Correct. g = -9.8ms^-2
    However, in a question I posted, I was told not to take g = -9.8ms^-2 but, rather, 9.8ms^-2 without the implication of the direction.
    Why question then is why?
    In calculating range projectile, should g be a vector g where if upwards is +, then downwards i -. So in utilizing the equation for range projectile, g should be -9.8. However, a poster told me to disregard the signs and take g as a magnitude. This doesn't make sense. Acceleration of gravity is acting on the object during its trajectory.
     
  5. Dec 31, 2013 #4

    Bandersnatch

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    As long as you use a reference frame in which the displacement "upwards" is in the positve direction, you have to use -g, since that's the direction towards which the gravity acts.

    As far as I can see from your posts in the other thread I think you're refering to, the problem is that you use the negative value for g twice, so to speak, and thus get the wrong result.

    The general equation for projectile motion is: [itex]r=\frac{at^2}{2}+V_0t+r_0[/itex], where r, a, and V are vectors.
    If you reduce the equation to one dimension(y; where the positive values are "up"), you can simply say that [itex]a_y=-g[/itex], you end up with [itex]y=\frac{-gt^2}{2}+V_{y0}t+y_0[/itex]

    You seem to be wrongly using [itex]r=\frac{-at^2}{2}+V_0t+r_0[/itex]
    and then substituting [itex]a_y=-g[/itex], which erroneously gives you [itex]y=\frac{gt^2}{2}+V_{y0}t+y_0[/itex]

    That's all there is to it.
     
  6. Dec 31, 2013 #5

    Bandersnatch

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    I guess something might've been unclear in the above post. I meant g to be equal to +9.81 in it. It's the magnitude of the gravitational acceleration. The direction is the minus sign you put into the equation when you substitute -g for ay.
     
  7. Dec 31, 2013 #6
    I was referring to this
    https://www.physicsforums.com/showthread.php?t=730711

    If my value for g = -9.8, I get a correct value but in the negative. The book has a positive value. This means that, and echoed by another poster, that g must be positive.
    The confusion is, why should g in this equation be a magnitude. It doesn't make an iota of sense.
     
  8. Jan 1, 2014 #7

    Bandersnatch

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    The value of g is not -9.8. It's 9.8. If you're asked to substitute the value of the scalar g into your equation, you substitute the value without any information about the direction of the vector(i.e., the '-' sign). If you see g in an equation, it's a safe bet the direction was already included at one time.
    If you were asked to substitute some value for the vector ##a_y## then you'd have to put the minus together with the magnitude(g), as you'd be asked to provide the full information.

    The equation you're asked to use: [itex]x = \frac{v_0^2}{g}sin{2Θ}[/itex]
    is derived from:
    [tex]y=\frac{-gt^2}{2}+V_{y_0}t+y_0[/tex]
    and
    [tex]x=V_{x_0}t+x_0[/tex]
    the minus sign was already included when we substituted ##a_y=-g## when we reduced the general vector equation to one dimension.
    Try solving the first one for t, then substitute the result to the second one, and you'll see what happened to the minus sign that was originally there.
     
  9. Jan 1, 2014 #8
    You are correct.
     
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