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Another cannonball question

  1. Mar 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A Cannonball is shot upward from the ground into the air at t=0 sec. With initial velocity of 50m/s. Its height above the ground in metres is given by s(t)=50t-4.9t^2 .
    ----What is the velocity of the cannonball when it is 100m above the ground on the way up?
    "says the book" Hint: set s(t) = 100

    2. Relevant equations
    initial velocity = 50 m/s

    3. The attempt at a solution
    I don't want the answer, I just want to know how it is worked out, because I know how to do it by trial and error.
    If I set s(t)=100 like the book says, then 100=50t-4.9t^2
    I can't solve this equation without knowing t, am I right?
    Anyhow, by trial and error, i figured out t=2.73, and I think velocity at 100m = 36.5m/s...
    But what is the proper way to do this without using trial and error like I did?

    ie.
    The question I should ask myself first is what?
    1. What is t at 100m?
    2. Velocity = ? at 100m?
     
  2. jcsd
  3. Mar 13, 2015 #2

    SammyS

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    Wrong. In fact, to find the time, t, at which the cannon ball is 100m above the ground, you need to solve this.

    It's a quadratic equation.
     
  4. Mar 13, 2015 #3

    Merlin3189

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    I think you deserve credit for solving by trial and improvement. I often do this myself when I can't see a simple analytic solution.

    As Sammy says, it is a quadratic equation, so it's just a maths issue to solve it.

    WikiP explains it here
     
  5. Mar 17, 2015 #4
    Thanks to all for the help.
    I actually figured this out very quickly after a relaxing weekend.
    Fresh eyes matter.
    On a side note, whoever moved this thread from the 'Pre-calculus Math' section and into 'Introductory Physics Homework' section, really bruised my ego!
     
  6. Mar 17, 2015 #5

    SteamKing

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    You'll get over it. :wink:
     
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