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Projectile motion Question

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data
    a ball is thrown from the floor at an angle of 45 degrees and with an initial velocity of sqrt(6gD) where D is the height of the room. How far does the ball travel horizontally before hitting the ceiling?

    2. Relevant equations

    y=tan(a)x-gx^2/(2v^2cos(a)^2)

    3. The attempt at a solution

    I let the height equal D

    D=tan45x-gx^2/(2sqrt(6gD)^2cos(45)^2)

    D=x-(x^2/6D) Solving this as a quadratic equation I (get 3+/-sqrt3)/2. But this answer makes no sense since it should be in terms of D. Where am I going wrong? thanks
     
    Last edited: Jul 1, 2011
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  3. Jul 1, 2011 #2

    Dick

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    Show us how you got the answer that doesn't involve D using the quadratic equation.
     
  4. Jul 1, 2011 #3
    0=-D+tan(45)x-gx^2/(2sqrt(6gD)^2cos(45)^2), tan45=2cos(45)^2=1 and g will cancel. that leaves this quad equation:

    -D+x-x^2/(6D)

    using the quad formula:

    -x +/-(sqrt(1-4(-d)(2/6D))/(4/3)

    which simplifies to 3+/-sqrt3)/2
     
  5. Jul 1, 2011 #4

    Dick

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    That's a little chaotic. The quad formula is (-b+/-sqrt(b^2-4ac))/(2a). What are a, b and c in that formula? None of them have an 'x' in it. But the (2a) part has a D in it, doesn't it?
     
  6. Jul 1, 2011 #5
    a=-(1/6D) b=1 c=-D
     
  7. Jul 1, 2011 #6

    Dick

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    Alright. So what's (-b+/-sqrt(b^2-4ac))/(2a)? I think it has a D in it. The D's cancel in the numerator, not in the denominator.
     
    Last edited: Jul 1, 2011
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