? asked in Science & MathematicsMathematics · 9 years ago

projectile motion involving air resistance?

Suppose a projectile is launched from a height, h. It can be assumed that the projectile is hit horizontally i.e. the initial vertical component of velocity is 0 while the initial horizontal component of velocity is v_1(0). Note that I will have a set of experimental values for h and v_1(0)

I have to come up with a general equation that can be used to find the range of this projectile for the different values of h and v_1(0).

The only tricky part is that we cannot ignore air resistance (drag). And we can assume that f{drag} = kv - i.e air resistance is a function of velocity and it is directly proportional to velocity.

How would I be able to use newtons laws of motion along with the differential equations used to model air resistance and come up with a general equation to find the range of the projectile??

thanks alot!


my mistake

f{drag} = -kv

1 Answer

  • hfshaw
    Lv 7
    9 years ago
    Best Answer

    You have to solve 2 different differential equations; one for the horizontal component of the motion, and one for the vertical component.

    Let y be the vertical coordinate, with the positive direction upwards. Let x be the horizontal coordinate, with the positive direction in the direction of the initial velocity. We want to know the range of the projectile; that is, the value of x when y = 0 (i.e., when the projectile hits the ground). To find this, we need to find expressions for both x and y as a function of time.

    Now let's look at the forces acting on the projectile once it is launched.

    In the vertical direction, there is the constant (assuming the projectile stays close to the surface of the earth) gravitational force directed downward. There is no component of the gravitational force that acts in the x direction. (We are neglecting the fact that the Earth is round -- for small distances, we can make believe the Earth is flat.) F_grav = -m*g, directed along the y direction .

    The only other force acting on the projectile is the frictional force, which you are told is equal to F_drag = -k*v (remember that both F and v are vectors). We can resolve this force into two components F_drag_x = -k*v_x = -k*dx/dt and F_drag_y = -k*v_y = -k*dy/dt, where v_x and v_y are the components of the velocity vector in the horizontal and vertical directions.

    The x and y components of the net force vector are then given by:

    F_x = -k*x'


    F_y = -m*g - k*y'

    Now apply Newton's second law, F = m*a, recognizing that a_x = x'' and a_y = y'', and where a_x and a_y are the horizontal and vertical components of the acceleration. "m", of course, is the mass of the projectile.

    F_x = m*a_x = m*x'' = -k*x'

    F_y = m*a_y = m*y'' = -m*g - k*y'

    So we end up with the differential equations:

    x'' + (k/m)x' = 0

    y'' + (k/m)y' = -g

    In addition, we have the initial conditions that x(0) = 0, y(0) = h, x'(0) = v1(0), and y'(0) = 0

    The DE for x is easily solved. Let x' = u, then x'' = u' and :

    u' = -(k/m)*u

    du/vu = -(k/m) dt

    ln(u/c) = -k*t/m

    where c is the constant of integration.

    u = v_x = dx/dt = c*exp(-k*t/m)

    Solve for c using the fact that v_x(0) = v1(0)

    dx/dt = v1(0)*exp(-k*t/m)

    Now integrate this again. This time, we have a separable equation:

    dx = v1*exp(-k*t/m) dt

    x(t) = -(m/k)*v1*exp(-k*t/m) + d

    where d is the constant of integration. We know that x(0) = 0 so:

    x(0) = 0 = d - m*v1/k

    d = m*v1/k


    x(t) = (m*v1/k)*(1 - exp(-k*t/m))

    This is the equation for the horizontal distance as a function of time.

    Now solve the differential equation for y.

    y'' + (k/m)y' = -g

    Again, let u = y', then:

    u' = -(g + (k/m)*u) = -(k/m)*(u + g*m/k)

    du/(u+g*m/k) = -(k/m) dt


    ln((u + g*m/k)/c) = -k*t/m

    u = v_y = y' = c*exp(-k*t/m) - g*m/k

    Initially, the vertical velocity is zero, so:

    0 = c - g*m/k

    c = g*m/k

    dy/dt = (g*m/k)*(exp(-k*t/m) - 1)

    Separate the variables and integrate again:

    dy = (g*m/k)*(exp(-k*t/m) - 1) dt

    y(t) = (g*m/k)*[(-m/k)*exp(-k*t/m) - t] + d, where d is the constant of integration.

    At time t = 0, y(0) = h, so:

    h = d - g*(m/k)^2

    d = g*(m/k)^2 + h


    y(t) = (g*m/k)*[(-m/k)*exp(-k*t/m) - t] + g*(m/k)^2 + h

    y(t) = (g*(m/k)^2)*[1 - exp(-k*t/m)] - g*m*t/k + h

    Unfortunately, we can't rearrange this solution to solve for t as a function of y, in closed form using elementary functions. You will have to solve for the time when y = 0 numerically or by trial and error (using, of course, the appropriate values for m, k, and h), then use that value of t to calculate the range of the projectile.

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