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2D Homework Help First Post

  1. Feb 9, 2014 #1
    Hello Guys,
    This is my first post here. I need help on my homework that I have been working on all evening. Basically I am given video data. 10 trials of a plastic ball in projectile motion. I have to use Logger Pro. Four things I need to find out for each trial:
    x-velocity
    y-velocity
    launch velocity from x data
    launch velocity from y data

    I analyzed the video already.
    x velocity will the slope of the linear fit equation. In this case velocity is constant and acceleration is zero.
    How do I find y velocity!?
    I believe launch velocity will be found out using Pythagoras theorem using the first x-velocity value and y velocity value.

    I am very confused about this. Can somebody show me the way?

    Thanks!
     
  2. jcsd
  3. Feb 9, 2014 #2
    What does "y-velocity" mean? Assuming Y denotes the vertical direction, the vertical component of velocity changes all the time, so at what time are you required to find it?
     
  4. Feb 9, 2014 #3

    CWatters

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    Science Advisor
    Homework Helper

    You will need to tell us a lot more about the experiment in the video. Otherwise we have little idea what these mean..

    x-velocity
    y-velocity
    launch velocity from x data
    launch velocity from y data

    For most projectile motion problems the horizontal velocity (x-velocity ???) is assumed to be constant (eg equal to the x launch velocity) but not always.

    The vertical velocity (y-velocity ???) usually varies with time as it's subject to acceleration due to gravity.

    One way to calculate the initial vertical velocity might be to measure the height achieved, then apply the standard equations of motion (eg SUVAT equations). However this might be the wrong approach if I've misunderstood the problem.
     
  5. Feb 9, 2014 #4
    I was thinking the same thing because the vertical velocity is constantly subject to change. However, I just noticed that each trial I am given as video data is a bit different from the previous one as the launch angle is changing. (0 degrees,10 degrees,20 degrees,30 degrees,40 degrees,50 degrees,60 degrees,70 degrees,80 degrees,90 degrees)

    Horizontal velocity will be pretty close to launch velocity in pretty much every trial. So these launch angles must be used to find y-velocity/vertical velocity?
     
  6. Feb 9, 2014 #5
    I figured that out myself. One question:

    Why horizontal and vertical velocities are almost the same? My vx=1.69 and vy=1.70

    Thanks
     
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