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Acceleration and maximum range

  1. Jan 26, 2013 #1
    How do you calculate the acceleration and the maximum range of a jump (in this case a frog,) when the maximum vertical hight is 0.28m and the distance of acceleration is 0.086 m? I also tried to use the 45 degree, and the gravitation constant 9,81 m/s^2 but cant get it right..
     
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  3. Jan 26, 2013 #2

    haruspex

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    Is the max vertical height that's given for a vertical jump? For a 45 degree jump? Something else?
     
  4. Jan 26, 2013 #3
    I dont have any more info, so I don´t know
     
  5. Jan 26, 2013 #4

    haruspex

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    Pls post the question as posed, word for word. You might be missing something in the interpretation.
     
  6. Jan 26, 2013 #5
    Task 2: Calculate the acceleration and maximum range of a jump in the case of a frog. The maximum vertical hight is 0.28m and the distance of acceleration is 0.086m. The gravity constant g=9.81 m/s^2

    This is the task i got (word for word), nothing less nothing more
     
  7. Jan 26, 2013 #6

    haruspex

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    Ok, I read that as meaning the height given is for a jump of max range. (It's not entirely clear, but I think that's a fair reading.)
    Suppose the take-off speed is u. What would the max range be? What height would be reached?
     
  8. Jan 26, 2013 #7
    could be, yes. my teacher don´t speak english, so he translates all the tasks in google translate ... Honestly, i have no clue what he is asking for.
    So, i dont know, i dont know i dont know, and i am sorry for not answering your questions.
    I just need a suggestion
     
  9. Jan 26, 2013 #8

    haruspex

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    You mentioned a 45 degree take-off. What made you suggest that?
     
  10. Jan 27, 2013 #9
    I read in a physicbook that the maximum distance of movement of an object in 2 dimensions, irrespective of wind resistance, the angle will always be equal to 45
    (as evidenced by sin2 * 45 = 1)
     
  11. Jan 27, 2013 #10

    haruspex

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    On a level surface, yes. (It's different going up or down an incline.)
    So assuming 45 degrees, and a take-off speed of u, what would the height and range be?
     
  12. Jan 27, 2013 #11
    I dont have any take-off speed. The height i guess would be 0.28, the range could be the distance (0.086) ?
     
  13. Jan 27, 2013 #12

    haruspex

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    I know that. I wrote, let the take-off speed be u, an unknown. If the take-off angle is 45 degrees, what is the vertical component of speed at take-off (as a function of u)? What kinematic equations do you have relating distance, constant acceleration and initial and final speeds? Using that, what height would be reached (as a function of u)? Knowing the height the frog does reach, what value does that give you for u?
    No, that's the distance through which the frog was accelerating during its take-off. After you've worked out the take-off speed, you can use it in the following ways:
    - find out how long the frog takes to land, and thus how far he jumps
    - find out what the acceleration must have been during the take-off period.
     
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