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Maximum height acceleration

  1. Oct 6, 2012 #1
    Hi everyone! If i know that acceleration is the derivative of the velocity, why when i throw an object in the air when the object reaches the maximum height its velocity is 0 m/s and its acceleration is still -9.8 m/s^2 ? I mean, the derivative of 0 m/s is not -g but still 0.

    Who can explain me this ?
     
  2. jcsd
  3. Oct 6, 2012 #2

    Vanadium 50

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    What is the derivative of y = 4x - 4? What is the derivative of that when x = 1?
     
  4. Oct 6, 2012 #3
    the derivative is 4 and when x=1 is still 4.
     
  5. Oct 6, 2012 #4

    Vanadium 50

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    Do you see your mistake now?
     
  6. Oct 6, 2012 #5
    Velocity in this case(magnitude) is not constant.Therefore you have to differentiate velocity as a function of time this and find its value at the required time.However,since velocity increases uniformly and ive is linear,its derivative will be constant
     
  7. Oct 9, 2012 #6
    Vanadium 50 -

    Clever - interesting how you explained what he did wrong
    without explicitly stating what he did wrong. That is a good
    strategy to employ when teaching others. Thank You.
     
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