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Find theta given range, height, and projectile velocity

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    If a trajectory is shot at angle theta at a fixed projectile velocity of 50 m/s, what angle(s) could it be shot so that it just makes it over a wall 30 meters high, 220 meters down the path.


    2. Relevant equations

    d=1/2at^2+vt
    sinx*v=vertical component
    cosx*v=horizontal component


    3. The attempt at a solution

    Time connects the horizontal and vertical distances, so my approach is to solve for time and then substitute.

    On the vertical end, we can say:

    Vo=50*sinx
    a=-9.8
    d=30m

    Horizontal:

    Vo=50*cosx
    Vf=50*cosx
    a=0
    d=220

    I solve for time on the horizontal time in terms of x.

    d=vt
    220=50cosx*t
    t=220/50cosx


    I can now use that t and plug it into the vertical t, ensuring that x is the only variable and solve for x.


    d=1/2at^2+Vot
    30=1/2(-9.8)(220/50cosx)^2+50sinx(220/50cosx)


    After setting that up, I am stuck. I have tried it multiple times, combining them and using LCD, etc. I end up with a cosx and a sinx and can't use trig identities or anything to isolate x. I can't find any way to solve for the angle of launch



    Thanks for the help!!
     
  2. jcsd
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