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Dynamics: solve both trajectories

  1. Oct 10, 2006 #1
    Hey all,

    Having trouble solving this one. Class is dynamics. 2D, constant accel.

    Shooting a cannon at a point above the initial. Inital velocity is 400m/s

    Cannon is at point A (origin) and we're shooting at point B @ (5000m, 1500m)

    I'm asked to find the two thetas that satisfy.

    So i got this so far, basicly the givens.

    In the x:

    ax = 0
    vx = v0*cos(Θ)
    x = v0*cos(Θ)*t

    In the y:

    ay = -g
    vy = v0*sin(Θ)-gt
    y = v0*sin(Θ)*t-(1/2)gt2

    I tried to eliminate t is the y position equation by

    1) solving x for t
    2) subing into y(Θ,t) for t to get y(Θ)
    3) solving for Θ

    I cant get through solving for Θ. Can I use the quadradic equation on y(Θ) to solve for Θ? How do i do that?
     
  2. jcsd
  3. Oct 10, 2006 #2

    radou

    User Avatar
    Homework Helper

    Yes, you should be able to solve the quadratic equation for [tex]\Theta[/tex], after plugging in the values of displacement 5000 m and 1500 m.
     
  4. Oct 10, 2006 #3
    Thats what my prof was saying, but im not sure how thats done? cause it some up to like


    -a*sec(Θ)^2 +b*tan(Θ) - c = 0

    so does it matter that Θ is in sec^2 for the a and tan for b?
     
  5. Oct 10, 2006 #4
    **In the voice of reason**

    Learn your trig identities backwards and forwards!

    *echo 7th grad algebra teachers voice

    **End voice**


    sec(Θ)^2 = 1 + tan(Θ)^2

    Sub and solve. Still an ugly problem, but I'm gald I figured it out.:tongue2:
     
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