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 and he ground. Write the equation of a parabola that models his path through the air, where x is horizontal distance from the start line in m and y is his height, in cm.

Answer 1

Answer:

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

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:

$y = a\cdot x^{2}+b\cdot x + c$

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