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Velocity/acceleration in calc via derivatives

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown vertically upward from the roof of a building 68.6 m high, with an initial velocity of 24.5m/s. Its height above the ground at time, t=>0, is given by

    h(t) = 68.6 + 24.5t-4.9t^2

    b) When does the ball reach it's maximum height?
    c) Find the maximum height reached by the ball.

    2. Relevant equations



    3. The attempt at a solution

    Well, i lent my friend the note i had for this and he went away on vacation, therefore i'm stuck with these 2 questions.

    Some notes: ball hits the ground in 7 seconds, when the velocity is -44.1m/s
    v(t) = -9.8t+24.5
    a(t)=-9.8
     
  2. jcsd
  3. Feb 20, 2009 #2

    Dick

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    The ball reaches it's maximum height when h'(t)=0. Right? Can you find t?
     
  4. Feb 20, 2009 #3
    2.5time, height 99.225
     
    Last edited: Feb 20, 2009
  5. Feb 20, 2009 #4

    Dick

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    Ok, that was fast. Not so bad a problem, yes?
     
  6. Feb 22, 2009 #5
    Yes, it was easy, but can I ask how you knew the ball reached its maximum height when velocity was 0?
     
  7. Feb 22, 2009 #6

    HallsofIvy

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    As long as the velocity is positive it is still going up and hasn't yet reached its highest point. When the velocity is negative is is already going down and has passed its highest point!
     
  8. Feb 22, 2009 #7
    So basically, when h'(t) = 0 , then h(t)'s y value will be at its highest point?
     
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