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Math Help - projectile launch velocities, given a position

  1. #1
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    projectile launch velocities, given a position

    struggling with this problem:
    using very simple projectile rules (no air resistance ect) and only gravity acting on the object, is it possible to:
    with a known: gravity, launch magnatude and coodinates which the partical passes through, to work out the launch angle (or x,y components?)

    using the position equations:
    X=Vx*t
    Y=Vy*t+(g/2)*t^2

    how do i find Vx and Vy?
    itld be easy if i knew time, but i dont, however i do know magnatude.
    sqrt(Vx^2+Vy^2)=M

    anyone got any ideas?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by smitzer View Post
    struggling with this problem:
    using very simple projectile rules (no air resistance ect) and only gravity acting on the object, is it possible to:
    with a known: gravity, launch magnatude and coodinates which the partical passes through, to work out the launch angle (or x,y components?)

    using the position equations:
    X=Vx*t
    Y=Vy*t+(g/2)*t^2

    how do i find Vx and Vy?
    itld be easy if i knew time, but i dont, however i do know magnatude.
    sqrt(Vx^2+Vy^2)=M

    anyone got any ideas?
    Vx = M~cos(\theta)
    and
    Vy = M~sin(\theta)
    where \theta is the launch angle.

    You'll have two equations in two unknowns which you can solve. My recommendation is to solve the x equation for cos(theta) and the y equation for sin(theta), then use
    sin^2(\theta) + cos^2(\theta) = 1
    to get an equation for t. (It's a biquadratic which you can solve for t^2 using the quadratic formula. Then solve for t from there.)

    FYI: You will get two angles.

    -Dan
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  3. #3
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    thanks dan,(thats my name also )
    so if:
    cos(\theta)=x/Mt
    and
    sin(\theta)=(y-(g/2)*t^2)/Mt
    i should be able to find t through:
    ((y-(g/2)*t^2)/Mt)^2+(x/Mt)^2=1
    is that right?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by smitzer View Post
    thanks dan,(thats my name also )
    so if:
    cos(\theta)=x/Mt
    and
    sin(\theta)=(y-(g/2)*t^2)/Mt
    i should be able to find t through:
    ((y-(g/2)*t^2)/Mt)^2+(x/Mt)^2=1
    is that right?
    Looks good to me.

    These are nasty, aren't they? I always used to scare my students by doing one of these as an example. Just to show them that I wasn't being mean by giving them the "hard" problems for homework. Really. That was the only reason.

    -Dan
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  5. #5
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    i got
    (g/2)*t^4-(M-g)*t^2+(x^2+y^2)=0
    but when i try some example numbers:
    g=0.3
    m=0.5
    x,y=10; ect
    my discriminant is always negative.

    i may have done that step wrong, i havent warmed up my maths brain in ages.


    again, thanks for any help
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by smitzer View Post
    i got
    (g/2)*t^4-(M-g)*t^2+(x^2+y^2)=0
    but when i try some example numbers:
    g=0.3
    m=0.5
    x,y=10; ect
    my discriminant is always negative.
    Two comments. First, g = 9.8 m/s^2 so you might as well use that value when cranking out the numbers.

    Also, I owe you an apology.
    Quote Originally Posted by smitzer View Post
    sin(\theta)=(y-(g/2)*t^2)/Mt
    This equation is not quite right. It should be
    sin(\theta)=(y+(g/2)*t^2)/Mt

    Sorry I didn't catch that earlier.

    -Dan
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  7. #7
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    solved this in the end
    i used the sin and cos to make tan,
    giving me a quadratic equation for it, with 4 answers

    thanks for the help, you certainly steered me in the right direction
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