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Math Help - Quadratic function - find max height, etc

  1. #1
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    Quadratic function - find max height, etc

    The height of a cannon ball, h metres, t seconds, after it has been shot out of a cannon is given by the formula h(t) = -12 - 20t - 5t^2

    a. Determine the maximum height of the cannon ball.

    b. How long does it take the cannon ball to reach its maximum height?

    c. How long does it take the cannon ball to reach the ground?

    d. How high off the ground is the cannon ball before it is shot out of the cannon?
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  2. #2
    A Plied Mathematician
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    Are you sure it's a -12 for the constant term? I ask because the answer to part d will be a little strange with a -12 there.
    What have you done so far to solve this problem?
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    Quote Originally Posted by Ackbeet View Post
    Are you sure it's a -12 for the constant term? I ask because the answer to part d will be a little strange with a -12 there.
    What have you done so far to solve this problem?
    That's the reason I'm asking in a forum. I got -12 for d too and I thought I was doing something wrong.

    I got:
    a. 8m
    b. 1.26 seconds
    c. 2.53 seconds
    d. -12 (??)
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  4. #4
    A Plied Mathematician
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    Ok. How did you get your answers? Justify them to me. (I'll agree with your d, by the way - that's a particularly straight-forward part of the problem!).
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    To determine maximum height I completed the square of the quadratic function, finding the vertex. The 'y' of the vertex being the height.
    To find b, I found the two x-intercepts and divided them by 2.
    For c, i multiplied the answer by 2.
    For d, I substituted x=0 into the equation.
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  6. #6
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    Quote Originally Posted by nadia95 View Post
    To find b, I found the two x-intercepts and divided them by 2.
    I'm not sure what you mean. If you divided the two x-intercepts by 2 you would get two numbers. But somehow you got a single answer of 1.26 seconds -- anyway this is wrong. I think you meant: ADD the two x-intercepts and divide the result by 2. If you do that, you will get -2 seconds.

    As it is, the original equation does not make sense, because the graph of the parabola is mostly to the left of the y-axis, implying negative time, which doesn't make sense. Can you double check to see if the original equation is right? This is what you wrote:
    h(t) = -12 - 20t - 5t^2
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    Sorry, I meant add the two x-intercepts and then divide the result by 2.
    But I still get 1.26 because the x-intercepts are x=-3.2649 and x=-0.7351.
    And this equation is from my online course. I'm positive this is what it says. I have emailed the teacher asking about it.
    I agree, even the original equation doesn't make sense.
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    If you take the average of two negative numbers, the result better be negative!
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    Quote Originally Posted by Ackbeet View Post
    If you take the average of two negative numbers, the result better be negative!
    Yeah but because I was finding a time, I figured it would be positive. So I changed it.
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  10. #10
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    Hmm. Unfortunately, that's not allowed mathematically. Here I come into a slight disagreement with eumyang. Negative time can make sense, especially in projectile motion problems. Time in these problems is arbitrarily assigned. In other words, I get to pick where my origin is, when I start my stopwatch.

    Let's do a thought experiment. Let's imagine a cannonball coming over the top of a wall, and let's say I start my clock right at the top when I see it coming over the wall to me (so I'm on the opposite side of the wall from where the cannonball was launched). Suppose I never saw the cannonball being launched. Now suppose I figure out the trajectory of the cannonball, and I want to find the time when the cannonball hits the ground. I guarantee you that you will find two solutions to the equation: one positive, and one negative - it's a quadratic in time, assuming you ignore wind resistance, the variation of gravity with height, the variation of atmospheric pressure with height, the Coriolis effect, and so on. The positive time that you find corresponds to when the cannonball hits the ground on your side of the wall. What's the negative time correspond to? It corresponds to the launch time! It's the time at which someone would have had to launch the cannonball in order to get to the top of the wall when you started your stopwatch and with the velocity you observed at your time t = 0.

    So you see, negative time can be quite significant, physically. Here are some things you won't find negative: mass and length.

    Does this all make sense?
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  11. #11
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    Thanks for all the help.
    The teacher replied and said it was a typo. -_-
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  12. #12
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    Well, I think you have what it takes to handle the new problem (you probably had that before!).

    Have a good one!
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