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Parametric equations problem

  1. Apr 1, 2009 #1
    The center field fence in a ballpark is 10 feet high and 400 feet from home plate. A baseball is hit at a point 3 feet above the ground. It leaves the bat at an angle of (theta) degrees with a horizantal at a speed of 100 miles per hour.

    x = (v0 cos(theta))t and y = h + (v0 sin(theta))t - 16t^2
    The initial velocity is v0 feet per second and the path of the projectile is modeled by the parametric equations. The projectile is launched at a height of "h" feet above the ground at an angle of (theta) with the horizontal.

    a) write a set of parametric equations for the path of the baseball.
    b) Use a graphing utility to graph the path of the baseball for theta = 15 degrees. Is the hit a home run?
    c) Use a graphing utility to graph the path of the baseball for theta = 23 degrees. Is the hit a home run?
    d) Find the minimum angle required for the hit to be a home run.

    the only part i'm having trouble with is part d). i don't know how to find the minimum angle required to hit a home run. i already calculated the v0 for this problem to be 146.67 ft/second. so i tried letting x = 400 since that's the distance needed to hit a home run. i got the equation 400 = (146.67 cos(theta))t and i solved for t and substituted back in the equation for y. then i got y = 3 + 400sin(theta) (1/cos(theta)) - 119(1/cos(theta))^2

    i then used the formula x = -b / 2a to find max/min for quadratic equations. i got (1/cos(theta)) = -400sin(theta) / -238. i rearranged and i got -238 = -400sin(theta)cos(theta). Then i have -238 = -200sin(2 theta). i divided by -200 and i took the arcsin of both sides. unfortunately i did not get the answer as -238/-200 isn't in the domain of the arcsin function. did i approach the problem the correct way? or is there a different way of doing it? please help.
     
  2. jcsd
  3. Apr 2, 2009 #2
    I believe that when the problem is requesting that you find the minimum angle required for a home run, it simply wants the angle that the ball is hit at to "just clear" the fence, i.e. hit the ball 400 ft horizontally and 10 feet vertically. If we simply plug in 40 for x, 10 for y, and and eliminate t, we can solve for the theta that meets these requirements.

    It is not asking for a maximum height or velocity, which is normally when we would be finding the max of a parabola and using -b/(2a) in this type of problem. I don't believe this is necessary here.

    Hint: In solving the resulting equation, you will need to know the trigonometric identity
    [tan(x)]^2 + 1 = [sec(x)]^2

    Since you are very focused on the method of solving the problem, I will tell you that I believe the answer is around 19.38 degrees. Try to see if you can get the same answer.
     
    Last edited: Apr 2, 2009
  4. Apr 3, 2009 #3
    ok i got the same answer as you now! the only part that bogged me down was forgetting that trig identity with tan and sec. thanks!
     
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