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Angle of a projectile

  1. Sep 2, 2011 #1
    "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?"

    Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
    using sinϑ=o/h, =0.5/8 = 3
    but it should be 21.6 .... do you know where I went wrong??
     
  2. jcsd
  3. Sep 2, 2011 #2

    cepheid

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    I see a couple of problems so far. The first is that I would interpret the 8 m/s as the horizontal component of the velocity. Therefore if you draw a right triangle consisting of the velocity vector and its two components, the 8 m/s is not the hypotenuse of the triangle, but rather the horizontal leg.

    The second problem is that you don't know what the vertical component is. You have to be careful here -- the 0.5 m provided is a distance, not a velocity. It's a little ambiguous what the "flight phase" of the jump is, but if you assume 0.5 m to be the maximum height that he reached, then you have enough info to solve the problem.
     
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