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ANOTHER parabola related question! ._.

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    a cannon launches a projectile. H, is the height in metres that the cannon ball is above the horizon by t seconds.

    h= -0.75t^2 + 16t + 3


    2. Relevant equations
    What is the maximum height reached by the cannonball?
    3. The attempt at a solution
    Ok so i tried completing the square:
    h= -(0.75t^2-16t-3)
    h= -(0.75^2-16t+64)+64+3 < (Is this right, that instead of -64 it is + because in the first brackets i was -64 really?)
    h= -(0.75t-8)^2+67

    Therefore the turning point is (8,67)?

    So would the highest point not be 67 metres? because when i graph it in my calculator it is 88.333 or something like that.

    Thanks again >_<
     
  2. jcsd
  3. Oct 2, 2009 #2

    Mark44

    Staff: Mentor

    What you want to do is to factor out the coefficient of your squared term and the first-degree term, like so:
    h= -(3/4)t2 + 16t + 3
    = -(3/4)[t2 - (64/3)t + ___] + 3
    = -(3/4)[t2 - (64/3)t + (322/9)] + 3 + (3/4)(322/9)
    and so on.
     
  4. Oct 2, 2009 #3

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Square (.75t-8). You don't get what you think you get. Try completing the square again, and keep in mind that the coefficient of t2 is not 1 here! You need to adjust for that
     
  5. Oct 2, 2009 #4
    oh shoop da woop thank you guys so much!

    After deciphering some of your maths jingo, i got it! Thanks guys i really appreciate your help!

    and Mark a special thanks to you for replying to all my threads so hastily!

    Thanks guys :D
     
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