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Kong Physics

  1. Mar 1, 2008 #1
    For a physics experiment, we had to take a kong (dog toy with a rubber ball on the end of a rope), twirl it and then throw it as far as we could. We were told to record how long it was in the air for, and the distance for two throws. This was all was we were told to record. Afterwards, we need to find the maximum height the kong reaches, it's horizontal velocity as it is in the air and it's vertical velocity when it hits the ground.

    Throw #1-

    Distance: 52m
    Time: 4.9s

    Throw #2-

    Distance: 65m
    Time: 3.84s

    So far, I have been able to figure out the horizontal velocity by using V=d/t.

    #1- 52m/4.9s=10.61m/s

    #2- 65m/3.84s=16.93m/s

    Now, I think that in order to find the vertical velocity, I would need to first find the kongs maximum height, and then divide this by half of the time, (since at the halfway point the kong will now be traveling horizontally, and would take the same amount of time to reach the ground as a kong that is dropped from the same height). However, I am not sure on how to find the maximum height.
  2. jcsd
  3. Mar 1, 2008 #2
    ok, here is what i got...


    Horizontal Velocity- 10.61m/s
    Vertical Velocity- 12.25m/s
    Maximum Height- 30.01m


    Horizontal Velocity- 16.93m/s
    Vertical Velocity- 9.6m/s
    Maximum Height- 18.43m
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