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Throwing a rock

  1. Aug 27, 2007 #1
    I don't know much about classical physics, if I throw a rock or mass M upward with an initial velocity Z, with gravity g and air resistance A acting against it, is this equation the right one ;

    [math]m\frac{dv}{dt} = (Z-g)m -Av[/math]

    Then if I want to know the maximal height, I just need to integrate v(t) to find the distance at time t. Right ?

    I'm really not sure about this equation, mostly because of the (Z-g) term. After all, it's about INITIAL velocity.
     
  2. jcsd
  3. Aug 27, 2007 #2

    mgb_phys

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    You have to define what you mean by air resistance 'A', is this a force or a drag coefficent, is it a fucntion of V or a constant?
    Generally the equation you want is s = ut + 1/2 gt^2

    See the formulae sticky thread for more equations.
     
  4. Aug 27, 2007 #3
    If you define air resistance to be a force proportional to the velocity

    Then the forces on it as it's going up are mg down and Av down, so you get

    [tex]\frac{dv}{dt} = -g - \frac{Av}{m}[/tex]

    You then have to solve this differential equation for v(t), using v(0) = Z as a boundary condition. Then you have all the information you need about the rock's movement.
     
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