Question

Two recording devices are set 3400 feet apart, with the device at point A to the west of the device at point B. At a point on a line between the devices, 400 feet from point B, a small amount of explosive is detonated. The recording devices record the time the sound reaches each one. How far directly north of site B should a second explosion be done so that the measured time difference recorded by the devices is the same as that for the first detonation?

Answer 1

The 2nd explosion should occur 690 feet North of B
Because;
Distance = Speed * Time . . And taking speed of sound to remain unchanged,
then Distance ∝ Time . . So Distance Differences ∝ Time Differences

E₁ to A = 2300 and E₁ to B = 300 Distance Difference = 2300 - 300 = 2000
Position E₂ for the same distance difference, AE₂ - BE₂ = 2000
Rearranging AE₂ = 2000 + BE₂. . Eq1.

As E₂ is at 90° to the AB line, triangle ABE₂ is a Right Angled Triangle,so
(AE₂)² = (AB)² + (BE₂)² . . . Given AB = 2600

Substitute for AE₂ from Eq1, and 2600 for AB,
( 2000 + BE₂.)² = 2600² + (BE₂)²
Reduces to BE = 690 feet = Answer

Checking answer; AE₂ = √ ( 2600² + 690² ) = 2690
So Distance Difference = 2690 - 690 = 2000
As expected for same Time Difference.

Answer 2

I hope this helped!