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Find the instantaneous velocity

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data
    a ball is thrown in the air. its height from the ground in metres after t seconds is modelled by h(t) = -5t^2 + 20t + 1. What is the instantaneous velocity of the ball at t=2 seconds?


    2. Relevant equations
    (f(a+h)-f(a))/h


    3. The attempt at a solution
    h(2) = -5(2)2 + 20(2) + 1 = 21
    h(2+h) = -5(2+h)2 + 20(2+h) + 1
    = -5(2+h)(2+h) + 40 + 20h + 1
    = -5(4+2h+2h+h2) + 20h + 41
    =-20 -10h – 10h-5h2 + 20h + 41
    = -5h2 + 21
    lim┬(h→0)⁡〖(f(2+h)-f(2))/h〗 = ((-5h^2+ 21)- 21)/h = (-5h^2)/h = -5h
    But then I'm supposed to let h = 0... so the instantaneous velocity is 0. This seems wrong to me? Unless because this is the maximum value, that is when the direction changes, so the velocity is 0?
    Thank you for helping!
     
  2. jcsd
  3. May 1, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    Yes, it is the maximum value and that's where the direction changes. 0 is correct.
     
  4. May 1, 2013 #3
    Great, thank you :)
     
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