A person standing on an open ground hears the sound of a jet aeroplane coming from north at an angle of 60 degrees with the ground level. But he finds the aeroplane right vertically above his position. If υ is the speed of sound, the speed of the plane is:

Let Vp be the velocity of the plane and Vt be the velocity of sound.
Let the time taken for the sound to reach the person be t.
The distance AB covered by the plane in time t is given by AB = Vp × t
The distance BC covered by the sound in time t is given by BC = Vt
In the right-angled triangle ABC, we have:
BC = Vt cos60° = AB
Vt cos60° = Vp × t
Since cos60° = 1/2, we have:
Vt (1/2) = Vp × t
Vt/2 = Vp × t
The vertical distance is given by AC = Vt sin60° = Vt√3/2
In the right-angled triangle ABC, we have BC = AB
Therefore, Vt/2 = Vp * t
Vp = Vt/2
The speed of the plane is Vp = υ/2