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What is the upward velocity?

  1. Sep 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity?

    2. Relevant equations



    3. The attempt at a solution

    Vv=Vsin[itex]\vartheta[/itex]

    So [itex]\vartheta[/itex] is the unkown

    Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
    using sin[itex]\vartheta[/itex]=o/h, =0.5/8 = 3

    Then VV=8Sin3 = 0.38 m/s (which is wrong, but where did I go wrong??
     
  2. jcsd
  3. Sep 1, 2011 #2
    Re: Projectile

    Is this the exact wording ? If it is then I don't think you have enough info to solve it.

    This does not make sense. You try to find an angle between a velocity and distance when you need an angle between velocities. you also assume forward velocity to be the total velocity not just the vertical component.
     
  4. Sep 1, 2011 #3

    lewando

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    Re: Projectile

    What is preventing you from using one of the standard kinematic equations? Hint: you know vy, ay, y, and y0.
     
  5. Sep 1, 2011 #4
    Re: Projectile

    Hi, yea the exact words are "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity? " I assumed because its asking for upward velocity = Vertical component which is Vsin(theata) so Theta is the unknown?

    The next questions after that asks "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?"

    So beacuse off this I assume I have the wrong idea regarding the first part; hence I need to not find the angle to calculate the upward velocity... if so Im confused...
     
  6. Sep 1, 2011 #5
    Re: Projectile

    oh thanks, so if I use 0=Vi^2 + 2ad

    0=Vi^2 + (2 x 9.81 X .05)
    Vi^2=9.81
    v=3.13

    Would I be correct??
     
  7. Sep 1, 2011 #6

    lewando

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    Re: Projectile

    Isolate the activity in the +y direction from the activity in the +x direction. Now ignore the activity in the +x direction. What's left is the jumper is jumping up 0.5m.
     
  8. Sep 1, 2011 #7

    lewando

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    Re: Projectile

    Yes, in m/s.
     
  9. Sep 1, 2011 #8
    Re: Projectile

    Hi again... the second part to the question "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??
     
  10. Sep 2, 2011 #9

    lewando

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    Re: Projectile

    bp_psy has pointed out that your vectors are not consistent. Since you now know Vy and Vx is given, work with them. If you want to use sinϑ=o/h, then you need to compute the magnitude of the Vx + Vy combined. Or you could just run with tanϑ=o/a.
     
    Last edited: Sep 2, 2011
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