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Kinematic equations for max height

  1. Jan 11, 2017 #1
    1. The problem statement, all variables and given/known data
    1.JPG
    2.JPG
    2. Relevant equations


    3. The attempt at a solution

    So (i) was easy enough and I got a time of 0.67 seconds.

    For (ii), since the horizontal time is 0.67 seconds this means the TOTAL time spent in air is also 0.67 seconds, so to calculate max height we split this time in half (since the ball has to go up to reach max height and then fall down again) and use the kinematic equations to find a value of S (distance)

    But they have kept the time, t, as 0.67 seconds and not 0.67/2, why is this?
     
  2. jcsd
  3. Jan 11, 2017 #2

    gneill

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    Staff: Mentor

    They didn't ask for the maximum height. They asked for the vertical displacement when the ball has traveled 1.5 m horizontally (i.e. at the horizontal distance of the ring).
     
  4. Jan 11, 2017 #3
    But max vertical displacement occurs when time spent in air is t/2 (compared to horizontal time) and when v=0, if you take a look at this video you can see the guy did the same thing; he's working out vertical displacement using v=0 which is the same as total air time divided by 2?
     
  5. Jan 11, 2017 #4

    gneill

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    Staff: Mentor

    But they did not ask for maximum displacement. They asked for the displacement at a particular horizontal distance.
     
  6. Jan 11, 2017 #5
    Ignoring for the moment the fact that they didn't ask you for the maximum height, your assumption that the maximum height occurs at t/2 is incorrect. That would only be true if it started and ended at the same height. If you actually did need to find the maximum height you would use the initial vertical speed and the known acceleration of gravity.
     
  7. Jan 11, 2017 #6

    CWatters

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    Science Advisor
    Homework Helper

    They keep the time at 0.67 seconds because that's how long it takes to get to a horizontal displacement of 1.5 meters.

    Aside: In general it's not necessary to split the motion into two phases (up and down) and apply the equations of motion twice. You can just apply the equations of motion once and answer will come out in the wash. Sometimes (not in this problem) you have to solve a quadratic and in that case you might get two answers (for example sometimes when calculating the time when a ball passes a certain height there are two solutions), in that case you may have to think about which is the required answer.
     
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