a rectangular object 25 m wide is to pass under a parabolic arc that has width of 32m at the base and a height 24m at the center. If the vertex of the parabola is at the top ofbthe arch, what maximum height should the rectangular object have?
Well, well, the rectangular object wants to pass under a parabolic arc... That's quite a daring move!
Now, to determine the maximum height of the rectangular object, we need to take a closer look at this parabolic arc. We know that the base width is 32m and the height at the center is 24m, with the vertex at the top.
Let's imagine the rectangular object going under this parabolic arc. Since we want the object to pass through without any trouble, the maximum height of the object should be equal to or less than the height of the parabolic arc at any given point.
Now, if the width of the rectangular object is 25m, and the base width of the arc is 32m, it means that the object will fit perfectly under the arc at its base.
However, as we move towards the center of the arc, the height increases. Therefore, the maximum height of the rectangular object should be less than the height of the parabolic arc at its center. In this case, it's 24m.
So, to ensure a safe passage, the maximum height of the rectangular object should be less than or equal to 24m, my friend!
To find the maximum height the rectangular object should have, we need to determine the highest point on the parabolic arc that the object needs to pass under.
Given:
- Width of the rectangular object: 25 m
- Width of the parabolic arc at the base: 32 m
- Height of the parabolic arc at the center: 24 m
Let's assume the origin (0,0) is at the vertex of the parabolic arc.
The equation of a parabola in vertex form is: y = a(x - h)^2 + k
Where (h, k) is the coordinate of the vertex.
Since the vertex is at the top of the arch, the k-coordinate of the vertex is the maximum height of the parabolic arc.
To find the equation of the parabolic arc, we need to find the value of 'a'.
We know that when x = 0 (at the base of the arc), y = 0 (at the vertex). So, plugging these values into the equation yields:
0 = a(0 - h)^2 + k
0 = a(-h)^2 + k
k = a(h)^2
Now we need to find the value of 'a' using the given information.
When x = 16 m (half the width of the base), y = 0 (at the vertex). So, plugging these values into the equation yields:
0 = a(16 - h)^2 + k
Substituting k = a(h)^2, we have:
0 = a(16 - h)^2 + a(h)^2
0 = 256a - 32ah + ah^2 + ah^2
0 = 2ah^2 - 32ah + 256a
Dividing throughout by 'a', we have:
0 = 2h^2 - 32h + 256
Now we can solve this quadratic equation for 'h' using the quadratic formula:
h = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -32, and c = 256. Substituting these values into the formula:
h = (-(-32) ± √((-32)^2 - 4(2)(256))) / (2(2))
h = (32 ± √(1024 - 2048)) / 4
h = (32 ± √(-1024)) / 4
Since we're looking for a real height, taking the square root of a negative number is not possible. Thus, there is no real solution to this equation.
Therefore, there is no maximum height that the rectangular object can have in order to pass under the given parabolic arc.
To determine the maximum height of the rectangular object to pass under the parabolic arc, we need to find the equation of the parabola and examine its properties.
1. Equation of the parabola:
Since the vertex is at the top of the arch, the equation takes the form y = ax^2 + h, where a is negative (since it opens downwards) and h is the height of the parabola at the center.
To find the value of a, we can use the given points on the parabola:
When x = 0 (center of the base), y = 24.
When x = 16 (half the width of the base), y = 0.
Using these points, we can set up a system of equations:
24 = a * 0^2 + h --> h = 24
0 = a * 16^2 + h
From the second equation, we can substitute the value of h:
0 = 256a + 24
Solving for a, we find that a = -24/256 = -3/32.
Therefore, the equation of the parabola is y = (-3/32)x^2 + 24.
2. Finding the maximum height of the rectangle:
Given that the rectangular object is 25 m wide, we need to find the maximum height such that its width does not exceed the width of the parabolic arc (32 m).
Let's consider the rightmost point of the rectangle, which is at x = 12.5 (half of the rectangle's width). Substituting this into the equation of the parabola, we can find the corresponding y-value:
y = (-3/32)(12.5)^2 + 24
Evaluating this expression, we can calculate the maximum height of the rectangle under the parabolic arc.
You know that the equation of the parabola is
y = 24-ax^2
Now plug in the point (16,0) to find a.
Now find y(12.5) to find the tallest object that will fit under the arch.
And of course, the vertex of the parabola will always be at the top of the arch!