Question

NO LINKS!! URGENT HELP PLEASE!!!

Cathy hits a golf ball down 250 yards down the fairway. If the ball reaches a maximum height of 25 yards above the ground, sketch and label a picture of the flight of the ball and write an equation that would give the height of the ball for any distance traveled. (Assume the golf ball travels in a parabolic path).

Answer 1

The equation is:

`\boxed{y=-0.0016(x-125)^2+25}`

where:

• x is the horizontal distance the ball travels (in yards).
• y is the height of the ball (in yards).

Step-by-step explanation:

The path of the golf ball can be modelled as a parabola that opens downwards, where x is the horizontal distance travelled (in yards) and y is the height of the ball (in yards). The equation of a parabola that opens downwards is a quadratic equation where the leading coefficient is negative.

If the horizontal distance (x) the golf ball travels is 250 yards, then its height (y) will be zero when x = 0 and x = 250. Therefore, the x-intercepts of the parabola are x = 0 and x = 250.

A parabola is symmetric about its axis of symmetry.

The axis of symmetry is the x-value of the vertex, and is midway between the x-intercepts. As the x-intercepts are x = 0 and x = 250, the axis of symmetry (and thus the x-value of the vertex) is x = 125.

If the maximum height the golf ball reaches is 25 yards, then the y-value of the vertex is y = 25. Therefore, the vertex of the parabola is (125, 25).

`\boxed{\begin{minipage}{5.6 cm}\underline{Vertex form of a quadratic equation}\\\\$y=a(x-h)^2+k$\\\\where:\\ \phantom{ww}$\bullet$ $(h,k)$ is the vertex. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}`

We can use the vertex form of a quadratic equation to write an equation for the height of the ball, y, for any distance travelled, x.

• As the parabola opens downwards, "a" is negative.
• The vertex is (125, 25) so h = 125 and k = 25.
• Since we know that the parabola passes through the origin (0, 0), x = 0 when y = 0.

Substitute these values into the formula and solve for a:

`\implies 0=-a(0-125)^2+25`

`\implies 0=-15625a+25`

`\implies -15625a=25`

`\implies a=-0.0016`

Therefore, the equation is:

`\boxed{y=-0.0016(x-125)^2+25}`

where:

• x is the horizontal distance the ball travels (in yards).
• y is the height of the ball (in yards).

Answer 2

The equation is...

• 
  `\text{y} = -(1)/(625)\text{x}(\text{x}-250)`
• Or it is
  `\text{y} = -(1)/(625)\text{x}^2+(2)/(5)\text{x}`
• Or it is
  `\text{y} = -(1)/(625)(\text{x}-125)^2+25`

Those listed forms are factored form, standard form, and vertex form in that order.

The graph is below.

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Step-by-step explanation:

• x = horizontal distance (in yards) the ball travels
• y = vertical distance (in yards) the ball travels

We'll have the ball start at the origin of the xy axis grid. This means (x,y) represents the location of the ball in the air. Assume the ground is completely flat, horizontal, and level. The ball moves to the right.

x = 0 and x = 250 are the two roots of this parabola. The ball starts on the ground at x = 0, goes up into the air, and comes back down when x = 250. This is to represent the 250 yard gap from start to finish.

• The root x = 0 means x is a factor of the quadratic function.
• The root x = 250 means x-250 is another factor of the function.

Together we have x(x-250) so far. It then leads to

y = ax(x-250)

where 'a' is some fixed constant. This constant tells how to vertically stretch the parabola, and it also tells us if the parabola opens up or down.

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At the midpoint of the roots 0 and 250 is (0+250)/2 = 125. This is when the ball reaches its peak height. This assumes there isn't any wind to slow down the ball or speed it up. Unfortunately wind will ruin any chance of symmetry we'll be going after; so that's why we assume there isn't wind to keep things relatively simple. We'll also ignore air resistance for the same reasoning.

x = 125 is the horizontal distance the ball traveled when it reaches the peak height of y = 25. In other words (x,y) = (125,25) is the vertex point of the parabola.

We'll use these coordinates to determine the value of 'a'.

y = ax(x-250)

25 = a*125(125-250)

25 = a*125(-125)

25 = a*(-15265)

a = 25/(-15265)

a = 25/(-625*25)

a = -1/625

The negative value of 'a' is expected because the parabola opens downward.

Therefore, we go from y = ax(x-250) to y = (-1/625)x(x-250)

That is the same as writing
`\text{y} = -(1)/(625)\text{x}(\text{x}-250)` which is one possible answer for the equation. We call it factored form.

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Optionally we could expand things out like so

y = (-1/625)x(x-250)

y = (-1/625)(x^2-250x)

y = (-1/625)(x^2)+(-1/625)(-250x)

y = (-1/625)x^2+(250/625)x

y = (-1/625)x^2+(2/5)x

That is the same as writing
`\text{y} = -(1)/(625)\text{x}^2+(2)/(5)\text{x}` which is another valid answer for the equation.

This is known as standard form y = ax^2+bx+c

• a = -1/625
• b = 2/5
• c = 0

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Lastly we could have used vertex form.

We calculated a = -1/625 earlier.

The vertex is (h,k) = (125,25) mentioned earlier as well.

We go from y = a(x-h)^2+k to y = (-1/625)(x-125)^2+25

That is the same as writing
`\text{y} = -(1)/(625)(\text{x}-125)^2+25` which is the third valid option for the equation.

The graph is below. I used GeoGebra to make the graph. Desmos is another good option.