u = 36 km h^$$-$$1 = 10 ms^$$-$$1,

$$v$$ = 320 ms^$$-$$1, $$\nu $$ = 8 kHz

The sound reflected from the building may be imagined to be coming from the mirror image. The driver is approaching the image-source which is also approaching him with the same speed. Hence the frequency of sound heard by the driver is

$$\nu $$' = $$\nu $$ $$\left( {{{v + u} \over {v - u}}} \right)$$

= 8 kHz $$\times$$ $$\left( {{{320 + 10} \over {320 - 10}}} \right)$$ = 8.5 kHz