Question

A long-jumper lifts off 3 m after starting his run, and lands 6 m later. When he is 8 m from the start line, he is 5 cm above the ground. Write the equation of a parabola that models his path through the air, where x is his horizontal distance from the start line in m and y is his height, in cm.
A. y = –x2 + 12x – 27
B. y = –3x2 + 4x – 9
C. y = 3x2 – 4x + 9
D. y = x2 – 12x + 27

Answer 1

Answer:

The equation of the parabola that models the path of the long jumper through the air is $y = -x^{2} +12$×$x-27$

Step-by-step explanation:

Mathematically, we know that parabolas are second-order polynomials and every second-order polynomials, also known as quadratic functions, can be constructed by knowing three different points of the curve. The standard form of the parabola is:

$y = a$ ×$x^{2} +b$×$+ C$

Where:

x - Horizontal distance from the start line, measured in meters.

y - Height of the long jumper, measured in meters.

a,b,c - Polynomial constants, measured in $\frac{1}{m}$, dimensionless and meters, respectively.

If we know that$(x_{1},y_{1}) = (3\,m, 0\,m), (x_{2},y_{2}) = (8\,m, 0.05\,m)$and $(x_{3}, y_{3}) = (9\,m, 0\,m),$ this system of linear equations is presented below:

$9\cdot a + 3\cdot b + c = 0 (Eq. 1)81\cdot a + 9\cdot b + c = 0 (Eq. 2)64\cdot a + 8\cdot b + c = 0.05 (Eq. 3)$

The coefficients of the polynomial are, respectively:

$a = -\frac{1}{100}, b = \frac{3}{25}, c = -\frac{27}{100}$

The equation of the parabola that models the path of the long jumper through the air is $y' = -\frac{1}{100}\cdot x^{2}+\frac{3}{25}\cdot x -\frac{27}{100}.$

But we need y measured in centimeters, then, we use the following conversion:

$y = 100\cdot y'$

Then, we get that:

$y = -x^{2}+12\cdot x -27$

Where x and y are measured in meters and centimeters, respectively.

The equation of the parabola that models the path of the long jumper through the air is $y = -x^{2}+12\cdot x -27.$