# Thread: Parabolic equation word problem

1. ## Parabolic equation word problem

Each cable of a suspension bridge is suspended (in the shape of a parabola) between two towers that are 400 feet apart and 50 feet above the roadway. The cables touch the roadway midway between the towers.
a) Find an equation for the parabolic shape of each cable. (I got this to as 800(y-0)=(x-0)^2. Feel free to help me double check if this is wrong.)
b) Find the length of the vertical supporting cable when x=100. (This is the one I don't understand. How do I solve this one? Do I just plug 100 in X or something?)

Edit- Another question I don't quite get.

Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of x feet, where the model for the path is given by y=-16/v^2(x)^2+s .

A bomber flying due east at 550 miles per hour at an altitude of 42000 feet releases a bomb. Determine how far the bomb travels horizontally before striking the ground.

I only got this far.
v=1.47 feet per second (converted from the 550 miles per hour)
s= 42,000 feet
How do I find the other stuff with only this information?

2. Originally Posted by C.C
Each cable of a suspension bridge is suspended (in the shape of a parabola) between two towers that are 400 feet apart and 50 feet above the roadway. The cables touch the roadway midway between the towers.
a) Find an equation for the parabolic shape of each cable. (I got this to as 800(y-0)=(x-0)^2. Feel free to help me double check if this is wrong.)
b) Find the length of the vertical supporting cable when x=100. (This is the one I don't understand. How do I solve this one? Do I just plug 100 in X or something?)

Edit- Another question I don't quite get.

Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of x feet, where the model for the path is given by y=-16/v^2(x)^2+s .

A bomber flying due east at 550 miles per hour at an altitude of 42000 feet releases a bomb. Determine how far the bomb travels horizontally before striking the ground.

I only got this far.
v=1.47 feet per second (converted from the 550 miles per hour)
s= 42,000 feet
How do I find the other stuff with only this information?
1. a) Assume that the first point you are given is at $\displaystyle x = 0$.

So you are given three points

$\displaystyle (0, 50), (400, 50), (200, 0)$.

Do you understand how these points are found?

Notice that your parabola can be modelled using the equation

$\displaystyle y = ax^2 + bx + c$.

Substituting the values of $\displaystyle x, y$ given, you get three equations in three unknowns.

$\displaystyle 50 = a(0)^2 + b(0) + c$
$\displaystyle 50 = a(400)^2 + b(400) + c$
$\displaystyle 0 = a(200)^2 + b(200) + c$.

Equation 1 gives $\displaystyle c = 50$, so

$\displaystyle 50 = 160000a + 400b + 50$
$\displaystyle 0 = 40000a + 200b + 50$

$\displaystyle 0 = 160000a + 400b$
$\displaystyle -50 = 40000a + 200b$

$\displaystyle 0 = 160000a + 400b$
$\displaystyle -100 = 80000a + 400b$

Subtract the second equation from the first to give

$\displaystyle 100 = 80000a \implies a = \frac{1}{800}$

Back substitution gives

$\displaystyle 0 = 200 + 400b$

$\displaystyle b = -\frac{1}{2}$.

So the equation of your parabola is

$\displaystyle y = \frac{1}{800}x^2 - \frac{1}{2}x + 50$.

1. b) Yes, let $\displaystyle x = 100$ and find what $\displaystyle y$ is.