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Missle Defense System (Projectile Motion)

  1. Sep 4, 2011 #1

    If it helps, the diagram for this problem can be found here:
    http://lon-capa.mines.edu/res/csm/csmphyslib/Mechanics/Kinematics/2D_Projectiles/MissileDefenseSystem.jpg [Broken]
    In the diagram, a2 is the acceleration a, as indicated by the equation below.

    You are designing a missile defense system that will shoot down incoming missiles that pass over a perimeter defense post. The set-up is shown below. An incoming missile passes directly above the defense base. Radar at the base can measure the height, h, and speed, v1, of the incoming missile. Your Patriot Rocket is set to fire at an angle of θ = 64.0 degrees from vertical. You design the Patriot Rocket so the magnitude of its acceleration is given by:

    a = A(e^−bt)

    where A can be set on your Patriot Rocket as it is fired, and b = 0.40 s-1. The direction of your Patriot Rocket's vector acceleration stays at the same angle, θ, for the entire trip. If an incoming missile passes over the defense base at a height of 4.90 km and at a constant speed of 729.0 m/s (this means that v1 is constant), solve for the value of A your Patriot Rocket must have in order to hit the incoming missile. You will also need to enter results from intermediate steps of your calculation, including the time ∆t in between launch and impact, and the horizontal distance ∆x from the launch station to the impact position.

    I thought you would have to substitute for t using equations for the xcomp and ycomp of displacement:

    ay(t) = Asin(64 degrees)*(e^(-.4t))
    vy(t) = integral of ay(t) from 0 s to t s = (Asin(64 degrees)/-.4)*e(-.4t)
    vy(t) = (Asin(64 degrees)/-.4)*(e^(-.4t)) + (Asin(64 degrees)/.4) = (Asin(64 degrees)/-.4)((e^(-.4t)) + 1)
    y(t) = integral of vy(t) = (Asin(64 degrees)/.4)*(((e^(-.4t))/.4) - 2.5 + t) = 4900m

    x(t) = (Acos(64 degrees)/.4)*(((e^(-.4t))/.4) - 2.5 + t) = 729t

    I substitute and I get (.4*729t*sin(64 degrees)) / (.4*cos(64 degrees)) = 729t*tan(64 degrees) = 4900

    6.72 = t*tan(64 degrees)

    I got 3.28 s. It's wrong.

    Am I doing this right at all? I'm scared...
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 5, 2011 #2


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    In the part I highlighted, what happened to the term that comes from the evaluating the integral at t = 0?
    Last edited by a moderator: May 5, 2017
  4. Sep 5, 2011 #3
    So it should be the integral of (Asin(64 degrees)/-.4)*(e^(-.4t)) + (Asin(64 degrees)/.4) for the acceleration? That's what I have in the line right below the one you highlighted. The line you highlighted was more like an indefinite integral. I know I probably shouldn't have indicated the limits if it was to be an indefinite integral.
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