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Quadratic Drag

  1. Jan 27, 2009 #1
    For a force F = −mg ± cv^2 (the sign is chosen so the drag force is opposite the motion), you can use a trick: let u = v^2, and generate a differential equation for u(x) (not u(t)!), which is not too hard to solve. (Hint: the chain rule gives this dv/dt=dv/dx * dx/dt should help.)
    a) Solve this to show that a projectile thrown vertically upwards from the ground with speed v0 reaches a maximum height y=V(t)^2/2g*(ln(1 + V(o)^2/V(t)^2, where V(t) is the terminal speed.
    b) Show that the projectile hits the ground with speed V(f) given by:
    V(f)^-2=V(o)^-2 + V(t)^-2

    So, part a) is ok, I have that after quite a bit of math, but I'm a little lost on b). When a projectile hits the ground, thats usually at t((y) = 0, but there is not time term in this equation.
     
  2. jcsd
  3. Jan 28, 2009 #2

    CompuChip

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    So let's call T the time when the projectile hits the ground.
    What do you know about the value of y(T)?
    The final velocity is V(T) = V(f).
    This should give you an equation.

    Also, please check the answer you are supposed to derive... is there really a V(t) in there?
     
  4. Jan 28, 2009 #3

    gabbagabbahey

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    I obtained the desired result, so I suspect that you may have used the wrong sign in front of 'c' during the falling portion of the motion.
     
  5. Jan 28, 2009 #4

    gabbagabbahey

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    I hope you aren't going to all the trouble of actually finding x(t) (or y(t) if that's what you're calling it)!

    All you really need to do is find [itex]u(x)=v^2(x)[/itex] as the projectile falls from a height of [tex]x=\frac{v_t^2}{2g}\ln\left(1+\frac{v_0^2}{v_t^2}\right)[/tex] to [itex]x=0[/itex]. The impact speed will simply be [itex]v_f=v(x=0)=\sqrt{u(x=0)}[/itex]
     
    Last edited: Jan 28, 2009
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