1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Differential equation: projectiles
  1. Differential Equations (Replies: 3)

  2. Differential equations (Replies: 5)

  3. Differential Equations (Replies: 1)

  4. Differential Equations (Replies: 4)