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Find angle (Projectile Motions)

  1. Dec 13, 2006 #1
    Alright so I thought I understood this well.

    I'll give an example of how I solve for range and such with the given information:

    A ball is thrown from a height of 1.86 m, at a speed of 21 m/s and at an angle of elevation of 59.4o. How far from the person who threw it does the ball land?

    Range: Vxt
    Vx: Vcos Theta
    Vy: V sin Theta

    The method in which I solve for t goes as follows:

    Vy = 21 * sin59.4

    Vy = 18.07



    Maximum Height: v²= 2a(s) +existing height of 1.86

    Maximum height: 18.07² /2(9.8) = s -> 16.66 + 1.86 = 18.52m





    Total time: Time down + Time Up
    s + 1.86 = 1/2at² + s = 1/2at²

    18.52*2 / 9.8 = t² + 16.66 * 2/9.8 = t²


    1.94 + 1.84 = 3.78s




    R = V cos Theta * t
    R = ~40.4 m



    The problem I encounter is simply that I do not know how, using this method, or by any other really, to compute the angle or initial velocity if Total time is not a given already.

    Say the equation was rearranged and said:

    A person threw the ball at an initial rate of 21 m/s from a height 1.86. It hit the ground at 40.4m. What was the angle of projection?





    Can anyone help show me how to work these backwards? Would be highly appreciated.
     
  2. jcsd
  3. Dec 14, 2006 #2
    v=21
    h=1.86
    (a means angle)
    Vx=V*cos(a)
    Vy=V*sin(a)
    when the ball hit ground ,x=40.4,y=0.
    put it into the equation andrevdh gave to you:
    Vx*t=x
    [tex]y = v_{yo}t - \frac{1}{2}gt^2[/tex]

    you may solve it,though it may difficult in maths.

    From now on,I advise you to learn more algbra.It is important in physics.
     
    Last edited: Dec 14, 2006
  4. Dec 14, 2006 #3

    andrevdh

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    For the given equation y = ....
    y will not be zero when the ball lands!
     
  5. Dec 14, 2006 #4
    Oh,Yes,I am so sorry!
    It should add h in.
     
  6. Dec 14, 2006 #5

    andrevdh

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    It seems you are trying to solve a though problem.
    See the section :"Angle θ required to hit coordinate (x,y)" in Wikipedia:

    http://en.wikipedia.org/wiki/Trajectory_of_a_projectile
     
  7. Dec 14, 2006 #6
    Like this y=h+Vy*t-(1/2)*g*t*t
     
  8. Dec 14, 2006 #7

    andrevdh

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    close, more like

    [tex]y = y_o + v_{yo}t - \frac{1}{2}gt^2[/tex]

    or if you take the launching point as (0,0) y = - 1.86
     
  9. Dec 14, 2006 #8
    Ah alright, thanks
     
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