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Trajectory of a projectile

  1. Feb 20, 2005 #1

    tony873004

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    A projectile fired from ground level at 25 m/s hits the ground 31.25 m away. At what angle was it fired?

    I wrote equations for t for both x and y

    t = delta x / velocity x
    t = 2 * viy/a

    and set them equal to each other to eliminate t
    and replaced velocityx with cos(theta)*25
    and replaced velicityinitialy with sin(theta)*25

    delta x / cos(theta)*25 = 2*sin(theta)*25/a

    25 / cos(theta)*25 = 2*sin(theta)*25/9.81

    sin(theta) * 25 * cos(theta)*25 = 153.28

    My trig is not that good. How do I continue? Or is this even the right way to tackle this proble? Is there an easier way?
     
  2. jcsd
  3. Feb 20, 2005 #2
    Assuming your previous work is correct, use the fact that sin(t)cos(t) = sin(2t)/2 to solve the last equation.
     
  4. Feb 20, 2005 #3

    tony873004

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    Thank you! That was exactly the trigonometry trickery I was looking for. :rofl:

    My problem now boils down to
    [tex]\frac{sin(2theta)25^2}{2} = 153.21825[/tex]

    [tex]2theta = sin^{-1} = 0.4905[/tex]

    [tex]2theta = 29.37[/tex]

    [tex]theta = 14.685[/tex]

    The back of the book, which rounds for significant figures list 15 degrees and 75 degrees. So I got one of the two answers with this method. But sin (75*2) also equal 0.5. And it makes sense that both the 15 and 75 degree angles should yield the same result in this type of projectile problem.

    So how do I make [tex]sin^{-1}[/tex] acknowledge that [tex]sin^{-1}0.5[/tex] equals both 30 (15*2) and 150 (75*2)?
     
  5. Feb 20, 2005 #4
    for your case, [tex]0 \leq \theta \leq 90\ degree[/tex], so
    [tex]0 \leq 2\theta \leq 180\ degree[/tex]
    so, you have to find all the values of theta that are in the above range and you have 2 values : 30 degree and 150 degree. so theta equals to 15 degree or 75 degree.
     
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