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Ball be thrown to have maximum horizontal distance

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown eastward across level ground. A wind blows horizontally to the east, and assume that the effect of the wind is to provide a constant force to the east, equal in magnitude to the weight of the ball. At what angle theta (to the horizxontal) should the ball be thrown to have maximum horizontal distance

    2. Relevant equations



    3. The attempt at a solution

    is this correct

    I believe the final vector for maximum distance should be v = i + j

    The vector for the wind is w = mass(ball)i

    So the vector of the ball being thrown pluss the wind vector, must equal the final vector v

    mass(ball)i + xi + yj = i + j

    x = 1-mass(ball), and y = 1

    thus the thrown vector is (1-mass(ball))i + y
     
  2. jcsd
  3. Sep 3, 2009 #2

    rl.bhat

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    Re: Vectors

    You have to deal this problem as the projectile motion problem.
    Only difference is the horizontal component of the velocity is not constant. The acceleration in this direction is also g.
    Find the time of flight and range.
    To find theta for maximum range, find the derivative of the range and equate it to zero.
     
  4. Sep 9, 2009 #3
    Re: Vectors

    hmmmmm never thought of it that way... still stuck though

    vox = vcos[tex]\theta[/tex] + mB, where mB = mass of the ball

    I do not understand how the horizontal acceleration is g, if i take that into account wouldnt that make the velocity vox = vcos[tex]\theta[/tex] + mBg. could you please elaborate on that statement
     
  5. Sep 9, 2009 #4

    rl.bhat

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    Re: Vectors

    The time of flight = 2vsinθ/g.
    Horizontal acceleration = mg.
    So the range R = vcosθ*2vsinθ/g + 1/2*mg*(2sinθ/g)^2
    Find dR/dθ and equate to zero to find θ for maximum range.
     
  6. Sep 12, 2009 #5
    Re: Vectors

    could u elaborate on how you got the time of flight

    x = [tex]\frac{2v^{2}cosQsinQ}{g}[/tex] +[tex]\frac{2mv^{2}sin^{2}Q}{g}[/tex]

    x = [tex]\frac{2v^{2}}{g}[/tex][cosQsinQ + msin2Q]

    dx/dQ = [tex]\frac{2v^{2}}{g}[/tex][2cos - 2QsinQ + 2msinQcosQ] + [cosQsinQ + msin2Q]

    how do i set that to 0, i could do it if that m was not there. did i make a mistake
     
  7. Sep 12, 2009 #6

    rl.bhat

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    Re: Vectors

    CosqsinQ = 1/2*sin2Q
    dx/dQ = 2v^2/g(cos2Q - msin2Q)
    Now equate it to zero and find Q.
     
  8. Sep 12, 2009 #7

    rl.bhat

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    Re: Vectors

    dR/dθ = 2v^2/g( cos2θ - msin2θ )
    Equate it to zero and find θ.
     
  9. Sep 13, 2009 #8
    Re: Vectors

    ok

    [tex]\frac{dx}{dQ}[/tex] = [tex]\frac{2v^{2}}{g}[/tex][cosQ + msin2Q]

    so i must get cosQ + msin2Q = 0


    I still dont see how to solve for Q when the m is in the equation...


    Also could you please elaberate on how you solved for t.....
     
  10. Sep 13, 2009 #9

    rl.bhat

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    Re: Vectors

    R/dθ = 2v^2/g( cos2θ - msin2θ ) = 0
    Or cos2θ = msin2
    Or sin2θ/cos2θ = 1/m
    so tan2θ = 1/m
     
  11. Sep 13, 2009 #10
    Re: Vectors

    so

    [tex]\theta[/tex] = [tex]\frac{arctan(1/m))}{2}[/tex]

    The problem states there is a numberical answer
     
  12. Sep 13, 2009 #11

    rl.bhat

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    Re: Vectors

    Sorry. I have edit this post.
    Horizontal acceleration is g, not mg. So the range is given by

    So the range R = vcosθ*2vsinθ/g - 1/2*g*(2vsinθ/g)^2
    =V2*sin2θ/g - 2v2sin2θ/g
    = v2/g[sin2θ - 2sin2θ]
    now find dR/dθ and equate it to zero.
     
  13. Sep 13, 2009 #12
    Re: Vectors

    x = v2/g[sin2Q + 2sin2Q]

    dx = v2/g[2cos2Q + 4sinQcosQ] = 0

    Q = 67.5
     
  14. Sep 13, 2009 #13

    rl.bhat

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    Re: Vectors

    It is not correct.
    2cos2Q - 2sin2Q = 0, because the force of wind opposes the motion.
     
  15. Sep 13, 2009 #14
    Re: Vectors

    The ball is thrown eastward... a wind blows horizontally to the east...

    doesnt that mean that the ball and the wind are in the same direction, not opposing

    also please explain your reasoning for find t, it is really buggin me
     
    Last edited: Sep 13, 2009
  16. Sep 13, 2009 #15

    rl.bhat

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    Re: Vectors

    I am sorry. I didn't notice that.
    While going up the particle experiences two accelerations. Vertical acceleration reduces the vertical component of the velocity. The horizontal acceleration has no effect on vy. So the time of will not change due to the horizontal acceleration.
     
    Last edited: Sep 13, 2009
  17. Sep 13, 2009 #16
    Re: Vectors

    ok so could u explain the t now...

    do you agree with my answer
     
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