I think all you need is the Pythagorean theorem. Am I missing something? Would you not have two intersecting circles?
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?
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.
LetCode:F * * | v(t+1) * | * | vt * | * | B * - - - - - - - - * A x
Let
and are feet apart.
The fireworks are at , due north of
In seconds, the sound travels feet from to
In seconds, the sound travels feet from to
Pythagorus: .
. . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . . . .
Hence: .
Then: .
Therefore, the fireworks display is about 9.25 miles north of