Question

suppose that two people standing 3 miles apart both see the burst from a firework display.after a period of time ,the first person standing at point A hears the burst. four seconds 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 ? Note that sound travels at 1100 feet per second

Answer 1

The fireworks display occur at 550.331 ft

The generic equation for a hyperbola provided by: is the best method for solving the problem in this instance.

$\frac{y^{2} }{b^{2} } - \frac{x^{2} }{a^{2} } = 1$ -

We presume that the y-axis is where the hyperbola's longer axis lies.

In this instance, the sound traveled a total of 1100 meters, and the value of a can be calculated as follows:

2a = 1100

a = 550ft

Given that the two people are 3 miles apart, we can calculate the value of c in feet as follows:  

1 mile = 5280 feet

∴ 3 miles = 15840 ft = c

And this is how we may determine what b is worth:

$b^{2} = c^{2} - a^{2}$

$b^{2} = 15840^{2} - 550^{2}$

$b^{2} = 250603100$

Our equation is thus given by:

$\frac{y^{2} }{250603100 } - \frac{x^{2} }{550^{2} } = 1$

And for this case we want to find the value of x when y = 3ft=15840ft and solving for x we get,

$\frac{15840^{2} }{250603100 } - \frac{x^{2} }{550^{2} } = 1$

$\frac{15840^{2} }{250603100 } - 1 = \frac{x^{2} }{550^{2} }$

$1.001- 1 = \frac{x^{2} }{550^{2} }$

$0.001 = \frac{x^{2} }{550^{2} }$

$302865.145= {x^{2}$

x = 550.331 ft.

The fireworks display occur at 550.331 ft.

Learn more about hyperbola here :

brainly.com/question/14307125

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