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Trajectory: integrating for position

  1. Sep 21, 2007 #1
    As usual I did this question I'm working on backwards, doing the section dealing with air friction first. Now realizing I must do the 'first part' and I think I am over thinking again.

    With no air resistance, I need the position of a trajectory from Newton's second law.
    I know gravity is the only force I'm considering, so I start with:

    -mg = ma right?

    My prof said when I integrate I'll get combos of sin and cos... I haven't been able to get that.. am I doing something wrong? What steps should I take to integrate this properly?
    Thanks!
     
  2. jcsd
  3. Sep 21, 2007 #2

    learningphysics

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    Yes, that looks right. What is the initial velocity though... that's probably where the sin and cos come in...
     
  4. Sep 21, 2007 #3

    Astronuc

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    If one applies Newton's law to a strictly vertical trajectory, then there is no angle, since the velcocity is parallel with the acceleration (force) of gravity.

    If one adds a horizontal component of velocity, that is where the angle (with respect to horizontal or vertical) enters the picture.

    Air resistance (friction) acts in both orientations (horizontal and vertical), and in opposition to the velocity of the object moving.

    See - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

    Look at freefall, then vertical launch, then horizontal launch.
     
  5. Sep 21, 2007 #4

    Doc Al

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    Sure: the acceleration of a projectile is g downwards.

    Perhaps he's talking about sin and cos of the launch angle.

    What's the exact problem?
     
  6. Sep 21, 2007 #5
    It's not simple to explain because the question refers to 'the projectile of 2.3'.

    2.3 compares the trajectory of a projectile subject to linear drag and one in a vacuum. Both have x and y components (not straight up), having the shape of someone hitting a baseball or firing a rocket at 45 degrees.

    Vo I assume would be zero then?

    Does that help?
     
  7. Sep 21, 2007 #6
    The problem is to get the position (x,y) of a projectile with no air resistance (from N2L I'm told) and then compare it to b.) which is considering linear air resistance.

    "Assuming there is no air resistance, write down the position (x,y) as a function of t, and eliminate t to give the trajectory y as a function of x."
     
  8. Sep 21, 2007 #7

    learningphysics

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    But if it is being fired... that implies that Vo isn't zero doesn't it? wouldn't horizontal initial velocity be Vocos45, and vertical initial velocity be Vosin45?
     
  9. Sep 21, 2007 #8
    Your right.
    45 isn't specified so I hope that's right... no angle is really given.
     
  10. Sep 21, 2007 #9

    learningphysics

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    oh... if it isn't specified... then maybe you're just supposed to go with an arbitrary angle [tex]\theta[/tex] ?
     
  11. Sep 21, 2007 #10
    But acceleration is what I'm starting with, so vcos45 and vsin45 is what I should receive through integration?
     
  12. Sep 21, 2007 #11
    maybe....
     
  13. Sep 21, 2007 #12

    learningphysics

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    Do the integration, if that is what is expected in the problem... in both the horizontal and vertical directions... introduce constants when required... then when solving for initial conditions (solving for the constants), that's where you'd put in the vcos(theta), vsin(theta)... don't assume 45 unless it is given.

    ie for the part without air resistance... horizontal force is 0... so [tex]a_x = 0[/tex]

    and [tex]a_y = -g[/tex]

    integrate to get vx, vy... then integrate again to get horizontal position, and vertical position.
     
  14. Sep 21, 2007 #13
    and then I would integrate those again for position?
     
  15. Sep 21, 2007 #14

    learningphysics

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    yes.
     
  16. Sep 21, 2007 #15
    F=ma
    -mg=ma
    integrating:
    -mg=m(vcosO +vsinO)
     
  17. Sep 21, 2007 #16

    learningphysics

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    careful... deal with the horizontal and vertical separately...

    start like this:

    ay = -g

    now integrate the left side to get vy, what do you get on the right side?
     
  18. Sep 21, 2007 #17
    hmm

    -mg = m(vsin45)
     
  19. Sep 21, 2007 #18

    learningphysics

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    I don't understand how you're getting that.

    you start here:

    [tex]-mg = ma_y[/tex]

    divide both sides by m

    [tex]a_y = -g[/tex]

    now integrate...
     
  20. Sep 21, 2007 #19
    [tex]v_{y}sin45 = -g[/tex] ?
     
  21. Sep 21, 2007 #20

    learningphysics

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    don't worry about the initial velocity for now... you want to integrate -g, with respect to time...

    suppose you had this question instead:

    [tex]\frac{dy}{dx} = -9.8[/tex]

    Find y.
     
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