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ikadub [295]
1 year ago
6

Suppose that two people standing 5 miles apart both see the burst from a fireworks display. After a period of time, the first pe

rson standing at point A hears the burst Three seconds later, the second person standing at point B hears
the burst. If the person at point Bis 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
Mathematics
1 answer:
RideAnS [48]1 year ago
5 0

Answer:

the fireworks display occured at x = 1,266,650 ft

Step-by-step explanation:

We are told that sound travels at 1100 feet per second

We will use the general equation for an hyperbola given by;

(y²/a²) - (x²/b²) = 1

We assume that the longer axis for the hyperbola is on the y axis.

In this problem, the total distance traveled by the sound is 1100 ft per second and we can find the value of "a" as:

2a = 1100

a = 1100/2 = 550

Also, since both people are 5 miles apart, we can find the value of c like this:

2c = 2(5)

So,c = 5 miles

Since the distance of sound was given in ft. Thus;

Lets convert 5 miles to ft.

1 mile = 5280 ft

Thus; 5 miles = 5 x 5280 = 26,400 ft

Now, we can find b² from;

b² = c² - a²

b² = 26400² - 550²

b² = 696960000 - 302500

b² = 696657500

Thus, plugging in the relevant values to obtain;

(y²/302500) - (x²/696657500) = 1

In this case we want to find the value of x when y = 5miles = 26400 ft

solving for x we have:

(26400²/302500) - (x²/696657500) = 1

This leads to;

2304 - (x²/696657500) = 1

Thus;

2304 - 1 = (x²/696657500)

2303 x 696657500 = x²

x² = 1604402222500

x = √1604402222500

x = 1,266,650 ft

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