Chapter 10.4, Problem 75AYU

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?

To determine

To find: Where the fireworks occurred.

Answer to Problem 75AYU

Fire display occurred at $50138\text{ feet }\approx 9.5\text{ miles}$ due north of the person at $\text{A}$.

Explanation of Solution

Given:

Two people are standing 2 miles apart both see the burst from a fireworks display. After a period of time the first person, standing at point $\text{A}$, hears the burst. One second later the second person, standing at point $\text{B}$, hears the burst. The person at point $\text{B}$ is due west of the person at point $\text{A}$, and the display is known to occur due north of the person at point $\text{A}$,

Formula used:

Equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $b^2=c^2-a^2$.

Calculation:

Let $\left(x,y\right)$ be location of firework and let it be a point north of $\text{A}$ such that time difference would be same as that for the first burst would. All such points form a hyperbola with $\text{A}$ and $\text{B}$ as foci and $\left(x,y\right)$ on it.

Let $x\text{-axis}$ contain $\text{AB}$ and let origin be midpoint of $\text{AB}$.

Sound travels at 1100 feet per second.

So person at $\text{A}$ is 1100 feet closer to the location than $\text{B}$.

Hence difference between $\left(x,y\right)$ to $\text{A}$ and $\left(x,y\right)$ to $\text{B}$ is a constant 1100.

Let the equation of the hyperbola be ,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$; $2a=1100$, $a=550$

Distance between $\text{A}$ and $\text{B}$ is $2c=2\text{ miles }=10, 560 \text{feet}$; $c=5280$.

To find $\text{b}$

$b^2=c^2-a^2={5280}^2-{550}^2=27575900$

Substituting $\text{a}$ and $\text{b}$, we get,

$\frac{x^2}{{550}^2}-\frac{y^2}{27575900}=1$

Let $x=5280$,

$\frac{{5280}^2}{{550}^2}-\frac{y^2}{27575900}=1$

Solving the above equation,

We get $y=50138$.

Hence, fire display occurred at $50138\text{ feet }\approx 9.5\text{ miles}$ due north of the person at $\text{A}$.