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Projectile motion with aerodynamic drag

  1. Oct 25, 2004 #1
    here's the problem:

    A projectile of mass "m" is launched from ground level at an angle of "theta" above the horizontal and with an initial speed "v". The flight of the projectile is resisted by an aerodynamic drag force modeled as uv^2, where "v" is the speed of the projectile and "u" is the drag coefficient.

    Find the maximum altitude "h" reached by the projectile and the horizontal distance "d" it travels before striking the ground.

    Let m=5 kg; theta=45 degrees; v=200 m/s; and u=100x10^-6 kg/m

    Compare answer to the case of no aerodynamic drag.

    any help would be great
     
  2. jcsd
  3. Oct 25, 2004 #2

    Pyrrhus

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    Homework Helper

    Use Newton's 2nd Law at Max height to determine its vertical acceleration.
    Remember Vy is 0 at max height, so there will be left only Vx at this point.

    [tex] \sum F_{y} = ma [/tex]
     
  4. Oct 27, 2004 #3
    i really don't know where to start with this problem but i do know that a computer program or excel must be used to solve it.
     
  5. Oct 28, 2004 #4
    The way I would approach this problem is such:
    (I wont put all the working, but hopefully will explain it enough)

    I would split the problem into components x and y.

    So:

    [tex] F_x = -uv_x^2 = ma = m\frac{dv_x}{dt}[/tex]

    and:

    [tex] F_y = -uv_y^2 - mg = ma = m\frac{dv_y}{dt} [/tex]

    you can solve these for [itex] v_x and v_y [/itex] in terms of t.

    Don't forget your integration constants!!
    (remember that at t=0

    [tex] v_x = v_0 cos(\theta) and v_y = v_0 sin(\theta)[/tex]

    You also know that :


    [tex] \frac{dx}{dt} = v_x and \frac{dy}{dt}= v_y [/tex]

    so you can solve for x and y by integration.

    as cycl said, v_y = 0 at max height -> you can then solve your v_y equation for t, and substitute this t value into your y equation to get the max height.

    For the max range, y = 0. Solve your y equation for t, then substitute that into your x equation.

    Alternatively you can find x in terms of y for the general case but
    the algebra can get bit nasty with this problem, linear drag is much nicer.

    Hope that helps,

    give us a yell if it doesn't

    Ty
     
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