1. ## Conic sections problem

Fireworks Display Suppose that two people standing 2 miles apart both see the burst from a fireworks display. After a period of time, the first person standing at point A hears the burst. One second later, the second person standing at point B hears the burst. If the person at point B is due west of the
person at point A and if the display is known to occur due north of the person at point A, where did the fireworks
display occur?

I tried using an ellipse, but I did not get anywhere. If the speed of sound is v=1129ft/s and the lightning is due north of A, then can't I use the fact that the sum of the distances from the foci is a constant? What am I doing wrong?

2. ## Re: Conic sections problem

I think all you need is the Pythagorean theorem. Am I missing something? Would you not have two intersecting circles?

3. ## Re: Conic sections problem

But if you use Pythagoras you get x=2miles, y=vt, z=v(t+1) and Z^2=X^2+Y^2 and when I sub in for x and y I get a negative time.

4. ## Re: Conic sections problem

I'm at a Y on a Smartphone at a disadvantage but l worked this out on Wolfram Alpha & l got a single + root ... get back to you.. Your hypotenuse is BF the distance you want is AF.. You want to use seconds and feet.

5. ## Re: Conic sections problem

l got 43.243 s and.9.24 mile.

6. ## Re: Conic sections problem

I know, you are right. I did not think of converting the distance to feet. I got it, thanks.

7. ## Re: Conic sections problem

Hello, christianwos!

You didn't convert the two miles to feet.
(Neither did I, the first time through.)

Fireworks Display
Suppose that two people standing 2 miles apart; both see the burst from a fireworks display.
After a period of time, the first person standing at point A hears the burst.
One second later, the second person standing at point B hears the burst.
If point B is due west of point A and if the display occurs due north at point A,
where did the fireworks display occur?

We don't need ellipses . . . We have a right triangle.

Code:
                        F
*
*  |
v(t+1)    *     |
*        | vt
*           |
*              |
B * - - - - - - - - * A
x
Let $\displaystyle v = 1129\text{ ft/sec}$
Let $\displaystyle x = 10,\!560\text{ ft}$

$\displaystyle A$ and $\displaystyle B$ are $\displaystyle x$ feet apart.
The fireworks are at $\displaystyle }F$, due north of $\displaystyle A.$

In $\displaystyle t$ seconds, the sound travels $\displaystyle vt$ feet from $\displaystyle F$ to $\displaystyle A.$
In $\displaystyle t+1$ seconds, the sound travels $\displaystyle v(t+1)$ feet from $\displaystyle F$ to $\displaystyle B.$

Pythagorus: .$\displaystyle (vt)^2 + x^2 \:=\:[v(t+1)]^2$

. . . . . . . . . . $\displaystyle v^2t^2 + x^2 \:=\:v^2t^2 + 2v^2t + v^2$

. . . . . . . . . . . . . $\displaystyle 2v^2t \:=\:x^2-v^2$

. . . . . . . . . . . . . . . $\displaystyle t \:=\:\frac{x^2-v^2}{2v^2}$

Hence: .$\displaystyle t \:=\:\frac{10,\!560^2 - 1129^2}{2(1129^2)} \:=\:43,24314023 \:\approx\:43.24\text{ sec}$

Then: .$\displaystyle F\!A \:=\:vt \:=\:1129(43.24) \:\approx\:48,\!817.96\text{ ft} \:\approx\: 9.25\text{ miles}$

Therefore, the fireworks display is about 9.25 miles north of $\displaystyle A.$

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### suppose that two peopl

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