Two persons stand facing each other, 200 m apart on one side of a high wall and at the same perpendicular distance to it. When one fires a pistol the other hears a report 0.60 s after the flash and a second 0.25 s after the first. Explain this and calculate:
(a) The velocity of sound in air.
(b) The perpendicular distance of the persons from the wall.
Let's first analyze the situation.
When the first person fires the pistol, the sound wave will travel directly towards the second person, who will hear the sound 0.60 seconds after seeing the flash. This delay in hearing the sound is due to the time taken for the sound wave to travel the distance of 200m.
Let's denote the velocity of sound in air as v, and the perpendicular distance of the persons from the wall as x.
Using the equation time = distance/velocity, we have:
0.60 = 200/v
Now, when the second person hears the sound a second time (0.25s after the first), this time another sound wave must have reflected off the wall and traveled the distance x twice. The total distance traveled by the second sound wave is 2x.
Using the equation time = distance/velocity, we have:
0.25 = 2x/v
Now, solving the two equations simultaneously:
0.60 = 200/v
0.25 = 2x/v
From the first equation, v = 200 / 0.60 = 333.33 m/s
Substitute the value of v in the second equation:
0.25 = 2x / 333.33
=> x = 0.25 * 333.33 / 2
=> x = 41.667 m
Therefore, the velocity of sound in air is 333.33 m/s, and the perpendicular distance of the persons from the wall is 41.667 m.