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Differential equation: projectiles

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A projectile of mass m is projected vertically upwards with speed U. In addition to its
    weight it experiences a resistive force mkv^2, where v is the speed at which the projectile is moving and k is a constant.
    Derive the equation of motion of the projectile:

    dv/dt = -(g + kv^2)

    Derive a differential equation for v as a function of z, where z is the height of the projectile above its starting position. Solve this equation to find the maximum height reached by the projectile.

    3. The attempt at a solution

    F = ma = m dv/dt

    -mg - mkv^2 = m dv/dt

    dv/dt = -g - kv^2 = -(g + kv^2)

    dv/dt = dv/dz * dz/dt = v dz/dt

    v dv/dz = -(g + kv^2)

    [v/(g+kv^2)] dv = - dz


    1/2k ln(g + kv^2) = -z + C

    ln(g + kv^2) = -2kz + 2kC

    When v = U, z = 0, 2kC = ln(g + kU^2)

    ln(g + kv^2) = -2kz + ln(g + kU^2)

    2kz = ln([g+kU^2]/[g+kv^2])

    Maximum z is when v = 0.
    z = 1/2k ln((g + kU^2)/(g)) = 1/2k ln(1 + kU^2/g)

    Is this correct?

  2. jcsd
  3. Mar 9, 2008 #2

    Shooting Star

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    Homework Helper

    Seems OK. Note that this is the eqn when the projectile is going up. When it's falling back, the eqn would be slightly different, because of the v^2 dependence of acceleration, unlike projectile motion without air resistance.
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