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Projectile Question

  1. Jan 2, 2006 #1
  2. jcsd
  3. Jan 2, 2006 #2

    Pyrrhus

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    Do you know why when there isn't air resistance the trajectory is parabolic? Do you know its proof?
     
  4. Jan 2, 2006 #3
    No. We just learned the basics - all we are supposed to need to know is that the experiments are taking place in a vacuum. This is the first question that has actually asked about air resistance.
     
  5. Jan 2, 2006 #4

    Pyrrhus

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    Well you should look up why the shape is parabolic when there isn't air friction, and then notice the effects of the air resistance by solving the linear 2nd order ODE by using the components in a plane motion (2d).
    They should have the following look
    [tex] x = C_{1} e^{-ct/m} + C_{2} [/tex]
    [tex] y = C_{3} e^{-ct/m} - \frac{mgt}{c} + C_{4} [/tex]
    The ODEs are
    [tex] -c \frac{dx}{dt} = m \frac{d^{2} x}{dt^{2}} [/tex]
    [tex] -mg - c \frac{dy}{dt} = m \frac{d^{2} y}{dt^{2}} [/tex]
     
    Last edited: Jan 2, 2006
  6. Jan 2, 2006 #5
    Hint 1: You will see the effect most clearly if your trajectory is as large as possible. Try to make it cover as much of the screen as you can. A good starting point might be 70 m/s and 65 degrees.

    Hint 2: Turn the trails on to make it easier to compare the case without air resistance to the one with.

    Hint 3: Use the fact that the air resistance is proportional to cv (v is the velocity of the projectile and c is a constant).

    Hint 4: Will air resistance have the same effect on the horizontal velocity as it will the vertical? Why is there a difference?

    That might help. I am lost as to what I am supposed to do. Your suggestion probably works but I don't know what a lot of the formula means. It is much more complicated than the way I am supposed to find the answer - which I don't know.

    I don't know what the constant is, and I still don't know how that will help me out.
     
  7. Jan 2, 2006 #6
    My attempt at an answer:

    The shape is still a parabola but it the maximum distance and maximum heights are both smaller. This change occurs because the air resistance force goes against the force of the projectile. The air resistance affecting the horizontal and vertical velocities is different because the constant is different in both cases.

    Thoughts?
     
  8. Jan 2, 2006 #7

    Pyrrhus

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    Yes, It's not a parabola, and what you mean by force of the projectile? There are only two forces acting on the projectile when it's shot, gravity and air drag. The constant (c) is the same for both x and y, the cvx and cvy are just components of the air drag force.

    No, yes there is. In our parametric equations, the [itex] a_{x} [/itex] is only affected by the air drag [itex] cv_{x} [/itex] component, on the other hand the [itex] a_{y} [/itex] is affected by the air drag component [itex] cv_{y} [/itex] and gravity.

    Remember, acceleration is the rate of change of velocity, and the speed is the magnitude or modulus of the velocity vector. It's true in some cases the speed is constant while the velocity is not, because of direction change. In this case the speed is not constant and neither the direction, as you can see because of the variable acceleration.

    Now about the range and max height, yes it'll be less than without air friction, but it'll be closer to the original when [itex] c \rightarrow 0 [/itex]. This means when c gets closer to 0.
     
    Last edited: Jan 2, 2006
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