A wing-tip Pitot tube at altitude 3 km reads 0.72 bar and ambient temperature is 268.66 K. Sea-level pressure, temperature and density are 1.01 bar, 288.16 K and 1.225 kg/m³. With g = 9.8 m/s², R = 287 J/(kg K) and standard atmosphere, find equivalent airspeed in m/s, to 2 decimals.
The tropospheric lapse rate from the stated data is $L=(288.16-268.66)/3000=0.0065$ K/m. Hydrostatic standard-atmosphere pressure is $p=p_0(T/T_0)^{g/(RL)}\approx69900$ Pa. For the intended low-speed Pitot relation, $q=p_t-p$ and $V_E=\sqrt{2q/\rho_0}$, giving approximately 58.59 m/s within the official 57.10–60.00 range. Equivalent airspeed uses sea-level density, not local density; using a compressible Pitot correction changes the result slightly but remains within that range.
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