Equation for Projectile Motion with Air Resistance?

Does anybody know the an exact equation for Projectile motion with air resistance? Bonus points to whoever knows how to graph it in Matlab as well

2 Answers

  • 1 decade ago
    Favorite Answer

    You can start with Newton's First Law

    F = ma

    and develop it into a differential equation

    F = m dv /dt

    Where F is the net force on the object, we will assume it has three dimensional motion.

    Fx = m dvx/dt

    Fy = m dvy/dt

    Fz = m dvz/dt

    Where x is into and out of the page, y is left and right, and z is up and down

    Thus in the z direction we must factor in gravity, drag will act in all directions, we will assume a spherical projectile. The y direction will only have the initial force and the drag force, the same in the x direction

    F0x - D = m dvx/dt

    F0y - D = m dvy/dt

    F0z - W - D = m dvz/dt

    The initial force is determined in your calculations it will be present if there is an initial acceleration, if there is no initial acceleration there will be no initial force.

    If we think conceptually about drag for a moment we can see that it relies on the square of the velocity and a variable value depending on the size of the projectile

    D is also based on the density of the fluid through which you are traveling, the orthographic projection of the area (area normal vector), and a constant value which is a scaling value called a drag coefficient.

    Since we assume a spherical projectile, D will be constant in all reference frames.

    F0 - D - Wz = m dv/dt Separate variables and integrate

    ((F0 - D - Wz) t) / m = v - v0

    D = (ρv^2AC)/2

    Wz = mg

    (F0/m - ((ρv^2AC)/2m) - g)t = v - v0 If you want the displacement we can use

    (F0/m - ((ρv^2AC)/2m) - g)t + v0 = dr / dt

    ((F0/m - ((ρv^2AC)/2m) - g)t + v0)dt = dr

    (F0/m - ((ρv^2AC)/2m) - g)(t^2/2) + v0t = r - r0

    • Login to reply the answers
  • 1 decade ago

    Try reading the Wikipedia article about drag:


    • Login to reply the answers
Still have questions? Get your answers by asking now.