How is the jet engine inlet danger area affected when airborne?
Let's assume the A320 is powered by V2500 engines, one of the options for it. This engine has an air mass flow of 355 kg/s. At sea level, air density is 1.225kg per cubic m. Hence, 355 kg/s = 355/1.225 = 290 cubic m/s.

Now, let's assume the aircraft is still at sea level, but now at Mach 0.8 (unrealistic, but we'll correct that next). The fan diameter is 1.6 m, which gives an intake area of 2.0 sq.m. Also, Mn 0.8 at 0ft is 272m/s. Hence, every second, the intake sweeps a volume of 272 x 2.0 = 544 cubic m. But, the engine only needs 290 cubic/m of air, per second assuming air density in the intake is the same as the air surrounding the aircraft. Therefore, the diameter of a streamline tube of the intake air will actually be smaller than the diameter of the engine inlet. It will be the area that sweeps 290 cubic m/s at 272m/s = 290/272 = 1.07 sq. m., or a diameter of 1.17m, not 1.6m.

Basically, the engine takes the airflow it wants, not what the intake area x forward speed provides. If the engine wants more (such as when the aircraft speed is low, or stationary, but the engine rpm is high, such as at the start of the takeoff roll), the engine will draw air from a large area in front of the engine (as per the max takeoff conditions diagram). Conversely, when the aircraft is at high speed, and the engine throttled back, the intake will spill the excess air it is providing (causing spillage drag).



Now, let's correct the fact the aircraft can't do Mn 0.8 at sea level. Let's re do the calculation at 35,000ft (10,700m). Here, air density is 0.38 kg per cubic m, and pressure and temperature are 3.46 psi and  219 Kelvin (-54C), compared with 14.7psi and 288 Kelvin (15C) at sea level. Thus our 355 kg/s, which is actually a corrected airflow, is a physical (real) air flow of 95.8 kg/s at 35,000ft since theta = 219/288 = 0.76 and delta = 3.46/14.7 = 0.235. Now 95.8 kg/s at 0.38 kg/ cubic m = 252 cubic m per second. Also, Mn 0.8 at 35,000ft is now 237m/s, not 272 as it was at seal level. Hence, every seond the intake area of 2 sq.m is sweeping through 2 x 237 = 474 cubic m per second. But we only want it to sweep 252, hence we need to find the diameter of a streamline tube that will cause this, at 237m/s. Therefore, we need an area of 252/237 = 1.06 sq. m., which occurs with a diameter of 1.16m. This compares with the physical inlet diameter of 1.6m.

Thus, at Mn 0.8, 35,000ft, max power, the green area at the engine intake is now a tube, of 1.16m dia, that extends in front of the aircraft. For how far this extends, depends as suggested, on what reaction time is required for the aircraft to manoeuvre away from an obstacle in this region, or for the object (a bird?) to manoeuvre out of the aircraft's way. 

The exact figure of 1.16 is questionable, as the assumption of air density in the intake not changing from the surrounding air is not entirely realistic. But the general result, is I believe, reasonable.