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Dynamics, Angle of projectile

  1. Apr 16, 2008 #1
    This problems has given me a lot of difficulty even asking several people. Kinematics of Particles using 2D vectors.

    v(initial) = 400 m/s. Determine two angles of elevation (theta) which will permit the projectile to hit the mountain target at B = (5 km)i+(1.5 km)j.

    wont directly let me host url because i'm new sorry. pm if picture needed thanks all.
  2. jcsd
  3. Apr 16, 2008 #2
  4. Apr 16, 2008 #3


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    Homework Helper

    Welcome to PF!

    Hi whtan20! Welcome to PF! :smile:

    With difficult questions, go one step at a time …

    First step: what equations do you know that might help here? :smile:
  5. Apr 16, 2008 #4
    I know that

    and integrating..v=dx/dt

    x = 400tcos(theta)
    y = 400tsin(theta)

    and from here i'm clueless as where to go. Thanks.
  6. Apr 16, 2008 #5
    x = v0*cos(θ)*t, so t = x/(v0*cos(θ))
    y = v0*sin(θ)*t - 1/2 g*t^2,
    y = x*tan(θ) - 1/2 g*x^2/(v0*cos(θ))^2
    use 1/(cos(θ))^2 = (sec(θ))^2 = 1 + (tan(θ))^2
    y = x*tan(θ) - 1/2 g*x^2/v0^2 - 1/2 g*x^2*(tan(θ))^2/v0^2
    then we get a quadratic equation for tan(θ) that is, (after plug all known values)
    765.63(tan(θ))^2 - 5000tan(θ) + 2265.63 = 0
    use quadratic formula and get
    tan(θ) = 6.0407, so θ = 80.6 degrees
    tan(θ) = 0.48987, so θ = 26.1 degrees
  7. Apr 16, 2008 #6
    Thanks a lot Simon, i did basically exactly that, but didn't know how to turn it into a quadratic, too many terms all over the place.
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