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Challenging Projectile Problem

  1. Mar 5, 2007 #1
    «Challenging» Projectile Problem

    1. The problem statement, all variables and given/known data
    The benches of a gallery in a baseball stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level 1 m above the ground and hits a mammoth home run. The ball starts at 35 m/s at an angle of 53 degrees with the horizontal The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit ?


    2. Relevant equations
    Equations of projectile motion
    [tex]R=\frac{u^{2}sin{2\Theta}}{g} [/tex]
    [tex]y=xtan\Theta(1-\frac{x}{R})[/tex]


    3. The solution i thought of
    Ok, I got a solution to this problem but it was very long and tedious (the original method which i thought of) I wonder if there could be some changes to the approach that would give the solution faster. A better method altogether is most welcome :)

    Here's my approach
    Let h be the height of the ball above the point where it was hit.
    First i calculated the range which is [itex]\approx 120 m[/itex]
    Then, differentiating eqn of path wrt time, i get-:

    [tex]\frac{dh}{dt}=\frac{d(xtan\Theta(1-\frac{x}{R}))}{dt} [/tex]

    [tex]\frac{dh}{dx}=tan\Theta(1-\frac{2x}{R})[/tex]

    [tex]dh=tan\Theta(1-\frac{x}{60}).dx [/tex]

    [tex]\int{dh}=tan\Theta\int{(1-\frac{x}{60}).dx} [/tex]

    [tex]h=\frac{4x}{3}(1-\frac{x}{120})[/tex]

    Another equation is from the bench thing, so height of bench [itex]y=x-109[/itex]
    Equating h and y, i get a quadratic,
    [tex]\frac{x^2}{90}-\frac{1}{3x}-109=0[/tex]
    The solution to this equation is [itex]\approx 115.3[/itex] , which tells that the bench will be 6th. (and matches the answer)
     
    Last edited: Mar 5, 2007
  2. jcsd
  3. Mar 5, 2007 #2
    Its not as elegant a soln but requires no calculus, just a messy quadratic:

    At 110m from the batter, draw a 45 degree line representing the slope of the benches. (not sure if the first bench is ground level, so may need slight adjustment)

    The point of intersection betwen the ball and this line can be described

    as Xf,Yf Xf=110+C and Yf=C (see note above re ground level)

    will hit the line at time t; hence

    t=(110+C)/Vx(init) and C=Vy(init)*t+1/2(g*t^2)

    subbing gives C=(110+C)tan(theta)+((110+C)/Vx)^2*(g/2),

    or 0=C(tan(53)-1)+440/3+.......ugliness
     
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