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Math Help - horizontal distance?

  1. #1
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    horizontal distance?

    The path of a thrown ball can be represented by the relation:

    h(d) = -(1/150)d^2 + (2/5)d + (3/2)

    where d metres is the horizontal distance travelled and h(d) metres is the height.

    (a) What is the maximum height reached by the ball?

    (b) What horizontal distance has the ball travelled when it reaches its maximum height?

    Answer key: (a) 7.5m (b) 30m

    Please help.
    Thanks.
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  2. #2
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    Hello, shenton!

    The path of a thrown ball can be represented by the relation:
    . . h(d) .= .-(1/150)d≤ + (2/5)d + (3/2)

    where d metres is the horizontal distance travelled and h(d) metres is the height.

    (a) What is the maximum height reached by the ball?

    (b) What horizontal distance has the ball travelled when it reaches its maximum height?

    Answer key: (a) 7.5m (b) 30m

    The height function is a quadratic; its graph is a down-opening parabola.
    . . It reaches its maximum at its vertex.

    The vertex of the parabola: y .= .ax≤ + bx + c
    . . is at: .x .= .-b/2a


    Our parabola has: a = (-1/150), b = (2/5)
    . . Hence: .d .= .-(2/5)/2(-1/150) .= .30


    (b) At maximum height, the ball has moved 30 m horizontally.


    (a) The maximum height is:
    . . h(30) .= .-(1/150)(30≤) + (2/5)(30) + 3/2 .= .15/2 .= .7.5 m

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  3. #3
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    Thanks Soroban for the detailed workings.

    The maximum height and horizontal distance now made sense.

    Maximum height is the y-value at its vertex and horizontal distance is x-value at its vertex.

    Two questions arises from here:

    (1) The x-value of the vertex can be calculated as:

    x = -b/2a
    = -(2/5)/2(-1/150)

    Entering -(2/5)/2(-1/150) in my calculator, I got 1/750. How did the answer arrive at 30?

    (2) I read somewhere that the y-value for the vertex can be calculated as

    y = c - (b/4a)

    Actually, in some other text, it would say the y-value can be calculated as

    y = - ( (b≤ - 4ac) / 4a )

    I don't know which is right.

    You listed a third method:

    y = ax≤ + bx + c

    The question is if the above 2 formulas are for calculating the y-value of the vertex, should I use your method or the above 2 methods?

    Thanks for the help.
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  4. #4
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    Hello Soroban

    For some reason, I could not post a reply to the "Quarters and Dimes" thread and another thread that I've posted.

    Thanks also for showing how to write the 2 equations in the Quarters and Dimes. I am familiar with the system of equations and by listing the two equations, the answer is more structured.

    I have difficulty coming up with the system of equations in another problem.

    I have to post this here as for some reason, I could not post a reply. Clicking reply will only give me a blank screen.

    The solution given by Captain Black does not list the two equations. Any chance, you could show the 2 equations.

    http://www.mathhelpforum.com/math-he...s-english.html

    Jack drives from her country home to the city, a distance of 380 km. He drives part of the way on back roads, at an average speed of 50 km/h, and part on highways, at an average speed of 100 km/h. The trip takes 5 hours. How far does he drive on each type of road?

    Answer key: 260 km on highways, 120 km on back roads

    Thanks.

    (Once again, I want to reply in the right thread but it would not let me reply - just a blank screen for the other two threads)
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  5. #5
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    Hello, shenton!

    (1) The x-value of the vertex can be calculated as:
    x \:= \:\frac{-b}{2a}\:=\:\frac{-\frac{2}{5}}{2\left(-\frac{1}{150}\right)}

    Entering -(2/5)/2(-1/150) in my calculator, I got 1/750.
    How did the answer arrive at 30?

    I'm sure you entered it incorrectly.

    Do some arithmetic first: . \left(-\frac{2}{5}\right) \div \left(\frac{-2}{150}\right) \:=\:\left(-\frac{2}{5}\right)\cdot\left(-\frac{150}{2}\right) \:=\:30



    (2) I read somewhere that the y-value for the vertex can be calculated as:
    . . y \:= \:c - \frac{b}{4a}

    Not quite . . . it is: . y \:=\:c - \frac{b^2}{4a}



    In some other text, it would say the y-value can be calculated as:
    . . y \:= \:-\frac{b≤ - 4ac}{4a}

    If you simplify this expression, it equals the one just above.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    All of the formulas are correct.
    I would use: .  x \,= \,\frac{-b}{2a} . . . and forget the rest.

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  6. #6
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    Thanks, Soroban for the clarifications.

    I worked according to your "manual" calculations and yes, it is 30.

    Seems to me that I have to be very cautious in entering the negative fractions in the calculator. I went to the extent of verifying the expression, -(2/5)/2(-1/150) in QuickMath Automatic Math Solutions (Algebra/Simplify) and the online calculator gave me 1/750 when I entered the expression -(2/5)/2(-1/150).

    Thanks also for explaining the formula for computing the y-value for the vertex. I will definitely forget the rest and use your recommendations.
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