JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Waves & Sound

The trick here is that what the man hears now was emitted earlier, when the plane was still behind him. In the time the plane flew from that emission point to the position directly overhead, the sound travelled along the slanting line that reaches his ear at $60^{\circ }$ to the ground. Both journeys take the same time $t$. The plane covers the horizontal distance $d$, so $d=v_{p}t$, while the sound covers a slant path of length $vt$ whose horizontal projection is that same $d$, giving $d=vt\cos 60^{\circ }$. Equating the two horizontal distances, $v_{p}t=vt\cos 60^{\circ }$, so $v_p=v\cos 60^{\circ }=\frac{v}{2}$. Written with the symbol used in the options, the plane's speed is $\frac{u}{2}$.