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Exciting distance problem! Only 2 variables are given.

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    An athlete executing a long jump leaves the ground at a 28° angle and travels 7.80 m. What was the takeoff speed?

    Xo = 0
    X = 7.80m
    Ay = -9.8m/s^2


    2. Relevant equations
    V=Vo+at

    X=Xo+VoT+(1/2)AT^2

    V^2=V^2o+2A(X-Xo)


    3. The attempt at a solution
    I made a right triangle with a 28° angle in the proper corner and established that the starting X velocity was H(cos(28)) and starting Y velocity was H(sin(28)) where H is the hypotenuse (initial speed). I've no idea what to do or if I'm on the right track. I feel like the problem should be giving me more stuff, and at the same time I can kind of see how the logic would work but I've no idea how to apply it.
     
    Last edited: Oct 7, 2011
  2. jcsd
  3. Oct 7, 2011 #2
    The y-component of the velocity should be X sin28. Since you have your component vectors, try to think of which one determines how long the athlete is in the air. That should get you set up.
     
  4. Oct 7, 2011 #3

    PeterO

    User Avatar
    Homework Helper

    Good start.

    Check this for some more hints.

    http://cnx.org/content/m13847/latest/

    And if you want to play with an applet

    http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/enapplet.html [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Oct 7, 2011 #4
    Yeah! Haha, typo, sorry! That's kind of what I was thinking but I'm unsure of how exactly to use what I have. Like I said, in my head I can see it all happening but I don't know where to start. Would I have to carry my hypotenuse variable all the way to the end of the problem and then do some sort of plugging in to get my final result?
     
  6. Oct 7, 2011 #5
    You have a known initial velocity in the y-direction, and you have a known acceleration in the y-direction. Both have an effect on the function of position versus time. You can use the kinematic equations (or integration, if that better suits your fancy) to determine how long the athlete is airborne.
     
    Last edited: Oct 7, 2011
  7. Oct 7, 2011 #6
    Thank you WJSwanson! I was lazy but you encouraged me to go through the formula! So this is what happened:

    I listed my known variables as follows:
    H = hypotenuse = initial velocity at 28° = final result

    x0 = 0
    x = 7.8 meters
    vx0 = vx = Hcos(28)
    a = 0

    t = ?

    y0 = 0
    y = 0
    vy0 = Hsin(28)
    vy = ?
    a = 9.8 m/s2

    I set up one of the kinematic formulas with what I knew about y motion:
    y = y0 + v0t + (1/2)at2
    0 = Hsin(28)t - 4.9t2
    -Hsin(28)t = -4.9t2

    (Hsin(28))/4.9 = t

    As you can see, my time still had an unknown variable (H) in it, and I could've plugged it right into the kinematic for X, but I didn't want to deal with any squares so I did the exact same thing I did with Y (kinematic equation with what I knew except for t), and got another t result. I equaled these two results and solved for H.

    Final answer: 9.60 m/s

    :smile:
     
  8. Oct 7, 2011 #7
    I'm getting a bit bleary-eyed at this point because it's a Friday night and the wife and I have been drinking, so I don't know that I'd be very reliable on checking your calculations. Your argument looks about right, and the dimensions came out correctly.

    Did you plug your derived values back into v0 = v <cos[itex]\theta[/itex] , sin[itex]\theta[/itex]>, i.e., 9.60m/s <cos28 , sin28> and work backward to check your answer? If you did (and you can use derivatives with respect to time to find the time t at which your athlete hits max height and then double it to find total transit time, etc.) and find total horizontal displacement? Because if it comes back as [itex]\Delta[/itex]x = 7.80m, then you're golden.
     
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