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Parametric Equations Word Problem

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a projectile launched at a height of h feet above the ground at an angle θ with the horizontal. If the initial velocity is v0 feet per second, the path of the projectile is modled by the parametric equations
    x=(v0cos θ)t and y=h + (v0 sin θ)t-16t2.

    The center-field fence in a ballpark is 10 feet high and 400 feet from home plate. The baseball is hit 4 feet above the ground. It leaves the bat at an angle of θ degrees with the horizontal at a speed of 100 miles per hour.

    Find the minimum angle required for the hit to be a home run


    2. Relevant equations



    3. The attempt at a solution

    So your basic equations are

    [tex]x=(146.67\cos\theta)t[/tex]
    [tex]y=3+(146.67\sin\theta)t-16t^2[/tex]

    by the question when x=400, y>10 the ball will pass over the fence

    so if I solve for the angle θ when x=400 and y=10 the angle I get should be the minimum passable.

    [tex]400=(146.67\cos\theta)t[/tex]
    [tex]10=3+(146.67\sin\theta)t-16t^2[/tex]

    [tex]\frac{400}{146.67\cos\theta}=t[/tex]

    sub that into the other equation

    [tex]y=3+(146.67\sin\theta)(\frac{400}{146.67\cos\theta})-16(\frac{400}{146.67\cos\theta})^2[/tex]

    I can get a common denominator and get it to

    [tex]7=400(146.67)^2\cos^2\theta\sin\theta-16(400)^2\cos\theta[/tex]

    but I'm not quite sure what to do after here to solve for θ.
     
    Last edited: Jul 20, 2010
  2. jcsd
  3. Jul 20, 2010 #2

    eumyang

    User Avatar
    Homework Helper

    Is it 3 or 4 in the 2nd equation? Your original problem says 4 feet.

    Assuming that it's 3:
    (I assume you meant to put a 10 in for y.) From here, try writing in terms of tan θ. I see a tan θ and a sec2 θ "hidden" in this equation, and you can use the pythagorean identity 1 + tan2 θ = sec2 θ. That way you'll end up with a quadratic in tan θ.


    69
     
  4. Jul 20, 2010 #3
    Yeah, its suppose to be 3. sorry.

    so I'd have

    [tex]
    10=3+(146.67\sin\theta)(\frac{400}{146.67\cos\theta})-16(\frac{400}{146.67\cos\theta})^2
    [/tex]

    [tex]7=(\frac{(400\sin\theta)}{\cos\theta})-(\frac{16(400)^2}{146.67^2\cos^2\theta})
    [/tex]

    [tex]
    7=400\tan\theta-\frac{16(400)^2\sec^2\theta}{146.67^2}
    [/tex]

    Then...

    [tex]
    7=400(146.67)^2\tan\theta-16(400)^2\sec\theta
    [/tex]

    [tex]
    0=-16(400)^2\tan^2\theta+400(146.67)^2\tan\theta-16(400)^2-7
    [/tex]

    Expand, factor out negative

    [tex]
    0=-(2560000\tan^2\theta-8604835.56\tan\theta+2559993)
    [/tex]

    Quadratic equation

    [tex]
    \tan\theta=\frac{8604835.56\pm\sqrt{-8604835.56^2-4(2560000)(2559993)}}{2(2560000)}
    [/tex]

    For the minus I get .3299 for the plus I get 3.031 the plus has way to high of an angle when I run an arctan on it and .3299 comes out as 18.27 degrees but this is wrong because I can check by plugging it back into the equation. and when x=400, y does not equal 10. It's close though. θ is suppose to equal "about 19.4 degrees" by the answer key.
     
  5. Jul 20, 2010 #4

    eumyang

    User Avatar
    Homework Helper

    You forgot to multiply the left side by 146.672 here. As it is, I wouldn't multiply both sides by 146.672 at all. I would simplify the coefficient of sec2 θ first:

    [tex]7 &= 400 \tan \theta - \frac{16(400)^2 \sec^2 \theta}{146.67^2}[/tex]

    [tex]7 &= 400 \tan \theta - \frac{14400}{121} \sec^2 \theta[/tex]

    (440/3 ≈ 146.67.)
    At this point, if you want, you could multiply both sides by 121 and then use the pythagorean identity.


    69
     
  6. Jul 20, 2010 #5
    Ok, now I see how to do it. Thanks!
     
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