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(D) Let $V_{P}$ be the speed of the plane and $\upsilon$ be the speed of sound.When the sound was emitted by the plane,it was at a position such that

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$\frac{\upsilon}{2}$

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(D) Let $V_{P}$ be the speed of the plane and $\upsilon$ be the speed of sound.When the sound was emitted by the plane,it was at a position such that the sound waves reached the observer at an angle of $60^{\circ}$ with the ground.Let the observer be at point $C$ and the plane be at point $A$ when the sound was emitted.The sound travels the distance $AC$ in time $t$,so $AC = \upsilon t$.The plane travels the horizontal distance $AB$ in the same time $t$,so $AB = V_{P} t$.From the geometry of the triangle $ABC$,where $\angle ACB = 60^{\circ}$ (angle of elevation),we have:$\cos(60^{\circ}) = \frac{BC}{AC}$ (This is not the correct approach based on the diagram).Looking at the diagram,the angle $60^{\circ}$ is given at $A$ and $C$. The horizontal distance covered by the plane is $AB$ and the sound travels distance $AC$.Using the relation $V_{P} = \upsilon \cos(60^{\circ})$:$V_{P} = \upsilon 	imes \frac{1}{2} = \frac{\upsilon}{2}$.

$A$ person standing on an open ground hears the sound of a jet aeroplane coming from the north at an angle of $60^{\circ}$ with the ground level. However,he finds the aeroplane right vertically above his position. If $\upsilon$ is the speed of sound,what is the speed of the plane?

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