It is desired to design a ski jump ramp. The ramp will be represented by a polynomial whose graph is a side view of the ramp. The ramp must fulfill the following specifications:
The ramp begins at a height of 100 feet and ends at a height of 10 feet above the ground.
From start to finish, the ramp covers a horizontal distance of 120 feet.
A skier will leave horizontally, and must fly at the end of the ramp at an angle of 30 ° above the horizontal.
1. Based on the specifications, determine the degree of the polynomial used, and determine the equations corresponding to the data.
I only need to know the degree of the polynomial so I can complete the rest of the homework
Thank You!
To determine the degree of the polynomial used to represent the ski jump ramp, we need to consider the number of turning points or bends in the ramp. Each bend represents an extreme point or a change in the direction of the graph.
Given the specifications, the ramp begins at a height of 100 feet and ends at a height of 10 feet above the ground. It covers a horizontal distance of 120 feet. We also know that the skier must leave the ramp at an angle of 30° above the horizontal.
Since the ramp starts at 100 feet and ends at 10 feet above the ground, there is a change in the height of 100 - 10 = 90 feet.
To find the degree of the polynomial, we need to determine if there are any bends or changes in the direction of the graph. In this case, there is only one bend at the end of the ramp where the skier leaves horizontally. Therefore, we can assume that the graph of the ramp is smooth with no additional bends.
A polynomial with one bend or turning point is represented by a quadratic equation, which has a degree of 2. So, to represent the ski jump ramp, you will need a polynomial of degree 2.
To find the equation corresponding to the data, we can start by considering the vertex form of a quadratic equation. The vertex form is given by:
y = a(x - h)^2 + k
Where (h, k) is the vertex of the parabola.
In this case, the vertex represents the highest point of the ramp, where the skier leaves. We also know that the horizontal distance covered by the ramp is 120 feet. Therefore, the x-coordinate of the vertex must be the midpoint between the start and end points, which is 120/2 = 60.
Since the ramp begins at a height of 100 feet and ends at a height of 10 feet above the ground, we can take the average of these heights to find the y-coordinate of the vertex. The average is (100 + 10)/2 = 55.
So, the vertex of the parabola representing the ski jump ramp is (60, 55).
Plugging these values into the vertex form equation, we get:
y = a(x - 60)^2 + 55
Now we need to determine the value of the coefficient "a" to complete the equation. This can be done by considering the angle at which the skier leaves the ramp.
Since the skier leaves at an angle of 30° above the horizontal, this angle corresponds to the angle that the tangent line at the vertex makes with the x-axis. The tangent line is perpendicular to the slope of the ramp at the vertex.
The slope of the tangent line can be determined by differentiating the equation of the ramp with respect to "x" and evaluating it at the x-coordinate of the vertex. The result will give us the slope of the tangent line.
Differentiating the equation, we get:
y' = 2a(x - 60)
Evaluating it at x = 60, we get:
y' = 2a(60 - 60) = 0
This means that the slope of the tangent line at the vertex is 0, and hence the tangent line is horizontal.
Since the tangent line is horizontal, it means that the ramp is at a constant height at the end. And since we want the ramp to end at a height of 10 feet above the ground, we can substitute this value into the equation and solve for "a":
10 = a(120 - 60)^2 + 55
Simplifying the equation, we have:
10 = a(60)^2 + 55
10 = 3600a + 55
3600a = -45
a = -45/3600
a = -1/80
Now, substituting this value of "a" back into the equation, we have:
y = (-1/80)(x - 60)^2 + 55
Therefore, the equation corresponding to the given specifications is:
y = (-1/80)(x - 60)^2 + 55
This equation represents the side view of the ski jump ramp as a polynomial of degree 2.
well, you only have two points, so a line would suffice to join them.
But, as the slope at the end of the ramp is not 90/120, that gives another constraint. That's like specifying a third point. So, use a quadratic to model the ramp.
y = ax^2+bx+c
y(0) = 100, so c=100
y(120) = 10, so 14400a+120b+100 = 10
y' = 2ax+b
y'(120) = 1/√3, so 240a+b = 1/√3
Now you can solve for a and b.