At an altitude of 4 km, a jet plane flew over the observer at a speed of 510 m / s.

At an altitude of 4 km, a jet plane flew over the observer at a speed of 510 m / s. How far from the observer will the plane be when the observer hears the sound?

Because the plane is supersonic, and flies faster than the speed of sound, then the observer will be the first to hear the sound emitted by the plane at the closest point from him (This is why, when flying over a supersonic plane, we hear a sharp pop, and only then a smooth descending hum).

t = h / 330 [m / s], where t is the time after which the observer hears the first sound
h – height
330 [m / s] – speed of sound in air

L = v * t, where t is the time after which the observer hears the first sound
v – aircraft speed
L is the distance that the plane will have time to fly

L = v * t = v * h / 330 [m / s]

L = 510 [m / s] * 4000 [m] / 330 [m / s] ~ = 6180 [m]

Answer: When the observer hears a sound, the plane will be at a distance of about 6180 m from him.