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Math Help - Quadratic Functions

  1. #1
    ADY
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    Quadratic Functions

    Hey guys,

    Maybe you could just check this for me and let me know if its correct, and help where i require it

    Ok so a ball is projected vertically from ground - modelled by:


    h=-5t^2 + 10t

    Interval, t = 0s to t=2.5s - plot this on your calc. setting window for the time interval.

    1, Find height of ball after 0.5s = 3.75m

    2, Find the 2 times when the ball is @ h = 0 - explain method = 2secs & ?? Help

    3, What is the maximum value of t which this
     f is a model for height of ball = explain

    4, Find the vertex of the Parabola and explain method = using

    y=ax^2 + bx + c (1) and ( -b\frac{b}{2a}, C - \frac{b^2}{4a} ) (2)

    Model
     h= -5t^2 +10t means using (1)

    a= -5, b= 10 and c= 0

    Vertex co-ordinates are:

    \frac{-b}{2a} = \frac{10}{2 x -5} = \frac{10}{-10} = -1 (For x Co-ord)

    C - \frac{b^2}{4a} = \frac{10^2}{4 x -5} = -5 (For y Co-ord)

    5, What is the velocity of the ball @ the vertex = Zero velocity

    6, Equation of Velocity-Time is given by
    V = 10 - 10t find the velocity of the ball @ t=0.5s & t=1.5s = 1m/s and -1m/s

    7, Explain what the plus and minus signs mean in Q6, what they indicate? = direction of travel

    8, How do you find the speed of the ball frm velocity? = the gradient, here it is
    1/(1.5-2) = -2m/s^-1 which is a speed of 2ms^1 in a downward direction

    9, By inserting h=1.8 , explain why solutions of quadratic:

    t^2 - 2t + 0.36 = 0

    gives the times at which the ball is at a height of h = 1.8 above the ground?

    Thanks for all your help in advance


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  2. #2
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    Quote Originally Posted by ADY View Post
    ...a ball is projected vertically from ground - modelled by: h=-5t^2 + 10t

    1, Find height of ball after 0.5s = 3.75m
    If you mean that h(0.5) = 3.75, then I agree with your answer.

    Quote Originally Posted by ADY View Post
    2, Find the 2 times when the ball is @ h = 0 - explain method = 2secs & ??
    Plug "zero" in for "h", and apply the Quadratic Formula.

    Quote Originally Posted by ADY View Post
    3, What is the maximum value of t which this  fis a model for height of ball = explain
    Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)

    Quote Originally Posted by ADY View Post
    4, Find the vertex of the Parabola and explain method = using y=ax^2 + bx + c (1) and ( -b\frac{b}{2a}, C - \frac{b^2}{4a} ) (2)

    Model  h= -5t^2 +10t means using (1)

    a= -5, b= 10 and c= 0

    Vertex co-ordinates are:

    \frac{-b}{2a} = \frac{10}{2 x -5} = \frac{10}{-10} = -1 (For x Co-ord)
    C - \frac{b^2}{4a} = \frac{10^2}{4 x -5} = -5 (For y Co-ord)
    I'm not sure what they mean by their first "method", unless they're wanting you to complete the square to find the vertex...?

    Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".

    Quote Originally Posted by ADY View Post
    5, What is the velocity of the ball @ the vertex = Zero velocity
    Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.

    Quote Originally Posted by ADY View Post
    6, Equation of Velocity-Time is given by V = 10 - 10t find the velocity of the ball @ t=0.5s & t=1.5s = 1m/s and -1m/s
    How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?

    Quote Originally Posted by ADY View Post
    7, Explain what the plus and minus signs mean in Q6, what they indicate? = direction of travel
    A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.

    Quote Originally Posted by ADY View Post
    8, How do you find the speed of the ball frm velocity? = the gradient, here it is 1/(1.5-2) = -2m/s^-1 which is a speed of 2ms^1 in a downward direction
    The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.

    Quote Originally Posted by ADY View Post
    9, By inserting h=1.8 , explain why solutions of quadratic: t^2 - 2t + 0.36 = 0 gives the times at which the ball is at a height of h = 1.8 above the ground?
    Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

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  3. #3
    ADY
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    Quote Originally Posted by stapel View Post
    If you mean that h(0.5) = 3.75, then I agree with your answer.
    No i mean 0.5 seconds. "Find the height of the ball after 0.5s" - Don't you make this 3.75 then?

    Quote Originally Posted by stapel View Post
    Plug "zero" in for "h", and apply the Quadratic Formula.
    Cant i just find the answer of the graph? In which case when h=0, the two times are 2secs and ?

    Quote Originally Posted by stapel View Post
    Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)
    How do i know where the ground is though?

    Quote Originally Posted by stapel View Post
    I'm not sure what they mean by their first "method", unless they're wanting you to complete the square to find the vertex...?

    Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".
    So are my co-ordinates wrong? - If so what do you make them, and could you please explain your method?

    Quote Originally Posted by stapel View Post
    Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.
    Woo!

    Quote Originally Posted by stapel View Post
    How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?
    Ok, i have no idea what i did, so we have v = 10 -10t - "Find the velocity of the ball @ t = 0.5s and t = 1.5s

    Quote Originally Posted by stapel View Post
    A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.


    Quote Originally Posted by stapel View Post
    The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.
    Is there a formula i could show this working?

    Quote Originally Posted by stapel View Post
    Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

    Ok so it looks like t^2 - 2t + 0.36 = 1.8 where do i go from here?

    Thanks for your help
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  4. #4
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    Quote Originally Posted by ADY
    Quote Originally Posted by stapel View Post
    If you mean that h(0.5) = 3.75, then I agree with your answer.
    No i mean 0.5 seconds. "Find the height of the ball after 0.5s" - Don't you make this 3.75 then? Mr F says: Stapel has agreed with you! She worded it like this because the way you expressed your answer was poor.

    Quote Originally Posted by stapel View Post
    Plug "zero" in for "h", and apply the Quadratic Formula.
    Cant i just find the answer of the graph? In which case when h=0, the two times are 2secs and ? Mr F says: You've been told how to do it ....

    Quote Originally Posted by stapel View Post
    Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)
    How do i know where the ground is though? Mr F says: It should be clear that the ground corresponds to h = 0.

    Quote Originally Posted by stapel View Post
    I'm not sure what they mean by their first "method", unless they're wanting you to complete the square to find the vertex...?

    Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".
    So are my co-ordinates wrong? - If so what do you make them, and could you please explain your method? Mr F says: Did you check your work like was was suggested?

    Quote Originally Posted by stapel View Post
    Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.
    Woo!

    Quote Originally Posted by stapel View Post
    How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?
    Ok, i have no idea what i did, so we have v = 10 -10t - "Find the velocity of the ball @ t = 0.5s and t = 1.5s Mr F says: Stapel is saying, kindly, that you have botched the basic arithmetic. Go back and do it right.

    Quote Originally Posted by stapel View Post
    A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.


    Quote Originally Posted by stapel View Post
    The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.
    Is there a formula i could show this working? Mr F says: Do what stapel said to do.

    Quote Originally Posted by stapel View Post
    Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

    Ok so it looks like t^2 - 2t + 0.36 = 1.8 where do i go from here? Mr F says: No! Where did you get this from? You certainly didn't follow the instruction very carefully.

    Substitute h = 1.8 into h = -5t^2 + 10t:

    -5t^2 + 10t = 1.8 => 5t^2 - 10t + 1.8 = 0.

    Now divide both sides by 5.

    Thanks for your help
    ..
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  5. #5
    ADY
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    Quote Originally Posted by mr fantastic;
    Mr F says: No! Where did you get this from? You certainly didn't follow the instruction very carefully.

    Substitute h = 1.8 into h = -5t^2 + 10t:

    -5t^2 + 10t = 1.8 => 5t^2 - 10t + 1.8 = 0.

    Now divide both sides by 5...
    You used the wrong equation, that's why i decided to start that other thread because this is not very well laid out,.

    By inserting h=1.8 , explain why solutions of quadratic:



    gives the times at which the ball is at a height of h = 1.8 above the ground?
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  6. #6
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    Quote Originally Posted by ADY View Post
    You used the wrong equation, that's why i decided to start that other thread because this is not very well laid out,.

    By inserting h=1.8 , explain why solutions of quadratic:



    gives the times at which the ball is at a height of h = 1.8 above the ground?
    No, I have used the correct equation.

    You have misundertsood the question. If you did the last small thing I left for you to do (divide through by 5) you would realise this.
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  7. #7
    ADY
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    I have and now i feel silly!

    Thank you for your help and guidence Mr.F!

    This one is solved. Apart from i dont know how to find the speed of the ball from the velocity?

    Thank you!
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