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Basic kinematics

  1. Jun 13, 2008 #1
    A cannon shoots at 400m/s towards a target 400m high and 15000m away from it, what angles must the cannon be a
    t?

    this is a kinematics question i have for homework and just dont know where to turn to. i tried with the equations but seem to always have something missing
     
  2. jcsd
  3. Jun 13, 2008 #2

    Kurdt

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    What have you tried? IF you show your work people will be in a better position to help you.
     
  4. Jun 13, 2008 #3
    i have tried everything i can using

    v(t)=vo+at
    x(t)=Xo+Vot+0.5at^2
    v^2-Vo^2=2a(delta)X
    or other variations of these equations

    every time i come up with 1 equation with 2 missing numbers and the i get stuck, for example, using the 1st and saying that at max height Vy0=0, i get stuck because i dont have sin(theta) or T
    using the 2nd and saying that Yf=400 i still dont have sin(theta) or T
    using the 2nd and saying Xf=15000 dont have cos(thetha) or t
    ising the 3rd dont have V unless i say Vy at max height is 0 but then dont have delta Y
     
  5. Jun 13, 2008 #4

    Kurdt

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    What you have to do is consider the x and y directions independently. This will give you 2 equations with 2 unknowns which you can then solve.
     
  6. Jun 13, 2008 #5
    still dont know how to do it, what am i supposed to do, i found that ymax height=8163 x sin(theta), dont see how that helps me,

    please help, i have another 4 questions of this sorty and cant continue my homework till i understand it, my next class is on monday and need to finish the work before it
     
  7. Jun 13, 2008 #6

    Kurdt

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    for the x direction what is the final distance going to be? What is the equation that will describe that? For the y direction what is the final distance? What is the equation that will describe that? you will now have two equations with two unknowns (time and angle) you can rearrange one of them to get time in terms of angle which you can then put into the second equation and solve for the angle which is what you want.
     
  8. Jun 13, 2008 #7
    i did that, i get , for x- t=3.75/cos(theta)
    then i must put that same t into my y equation, but how do i get from cos(theta) to sin(theta)??

    do you understand why i cant solve this or am i just being an idiot
     
  9. Jun 13, 2008 #8

    Kurdt

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    If you write down exactly what you get it would help a lot. From there I would try and make the equation a quadratic in tan (theta). You already should have a sin/cos term and a 1/cos^2 term. You can use trig identities to get them into tan from there.
     
  10. Jun 13, 2008 #9

    tiny-tim

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    Hi devanlevin! :smile:

    (Don't you mean 37.5?)

    Show us your y equation. :smile:
     
  11. Jun 13, 2008 #10
    for x
    15000=400xcos(th)xt
    t=37.5/cos(th)
    cos(th)=37.5/t

    for y
    400=400xsin(th)xt-4.9t^2
    t=81.633sin(th)
    sin(th)=0.01225t

    now from this how do i find theta??
     
  12. Jun 13, 2008 #11

    Kurdt

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    Not sure what all the bold bit is. You want to substitute the t in your y equation for the t you got from the x equation.
     
  13. Jun 13, 2008 #12

    tiny-tim

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    Hi devanlevin! :smile:

    (have a theta: θ :smile:
    and a squared: ² :wink:
    oh … and stop using x to mean "times")


    I don't know how you got the 81.633 either. :confused:

    Hint: get cosθ and sinθ on the left, then square and add. :smile:
     
  14. Jun 13, 2008 #13
    kurdt, the bold doesnt really matter,


    my problem is that if i substitute the t i found from the X, i get an equation with sin(theta) and cos(theta) which i think i can turn into tan(???) but i also get cos^2 from the t^2 and then dont know how to bring them to a result for theta

    400=400sin(th)x37.5/cos(th)-4.9x(37.5/cos(th))^2
     
  15. Jun 13, 2008 #14

    Kurdt

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