Question

suppose that two people standing 4 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. Two seconds later, the second person standing at point a 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 ft./s. The fireworks display is ____ feet north of the person at point A.

Answer 1

Explanation:

x = distance event to point A in ft

y = distance event to point B in ft = x + 2×1100

the distance is 2 seconds the speed of sound longer than x.

AB = 4 miles = 21120 ft (5280 ft in 1 mile)

so, we have a Pythagoras right-angled triangle situation :

c² = a² + b²

with c being the Hypotenuse (side opposite of the 90° angle), which is y in our case, as the right angle is at point A (the event is due north of A).

so, we have

y² = x² + 21120²

(x + 2200)² = x² + 21120²

x² + 4400x + 4,840,000 = x² + 446,054,400

4400x = 441,214,400

x = 100,276 ft = 18.99166667... miles

the fireworks display is 100,276 ft north of point A.

which is very far. I don't think they would see or hear anything.

so, I hope the information in the question is typed correctly. there are clearly some mistakes regarding point A and B. but I thought I could correct this logically.