1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook