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Projectile Motion of broad jumper

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A broad jumper begins his jump at an angle of 22 degrees from the horizontal. At the middle of the jump he reaches a height of 70.0cm
    a. What is his horizontal velocity
    b. What is his horizontal displacement
    ANSWER: 9.16m/s


    2. Relevant equations
    dh=vht
    dv=1/2at^2
    t=vv/a

    3. The attempt at a solution
    Sin 22 = x/70 x= 26.2
    26.2=1/2(9.8)t^2
    √t
    t= 2.3sec

    Since it's half way up, you'd have to double the time to find the entire time(?) needed for the entire horizontal displacement, but I have no clue what to do after getting time (if that's even right).
     
  2. jcsd
  3. Oct 26, 2012 #2

    haruspex

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    No, the angle is his initial trajectory. It does not provide a straight line to his highest point.
    What vertical component of speed would he need at take-off to reach that height?
    What, therefore, was his horizontal speed?
    Btw, it's not clear to me whether "horizontal displacement" refers to where he lands or at the highest point. I think it's the highest point.
     
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