The frequency of a car horn encountered a change from 400 Hz to 500 Hz. When the car approaches a vertical wall. If the speed of sound is 330 m s$$^{-1}$$. Then the speed of car is ___ km h$$^{-1}$$.

Let the speed of the car be $$v$$ m/s and the speed of sound be $$v_s = 330$$ m/s. The car horn has a natural frequency $$f_0 = 400$$ Hz.

When the car approaches a stationary vertical wall, the wall acts as a stationary observer first and then as a stationary source. From the perspective of the wall (observer), the frequency of sound received is $$f_1 = f_0 \frac{v_s}{v_s - v}$$.

The wall then reflects this frequency. Now the car acts as an observer moving towards the wall (the secondary source), so the frequency heard by the driver from the reflection is:

$$f' = f_1 \cdot \frac{v_s + v}{v_s} = f_0 \cdot \frac{v_s + v}{v_s - v}$$

This reflected frequency is $$f' = 500$$ Hz. Substituting:

$$\frac{500}{400} = \frac{330 + v}{330 - v}$$ $$\frac{5}{4} = \frac{330 + v}{330 - v}$$ $$5(330 - v) = 4(330 + v)$$ $$1650 - 5v = 1320 + 4v$$ $$330 = 9v$$ $$v = \frac{330}{9} = \frac{110}{3} \text{ m/s}$$

Converting to km/h: $$v = \frac{110}{3} \times \frac{3600}{1000} = \frac{110 \times 3600}{3000} = \frac{110 \times 6}{5} = 132$$ km h$$^{-1}$$.

The speed of the car is $$\boxed{132}$$ km h$$^{-1}$$.