Question

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still maintain the Mach number (0.5), how much faster must the jetliner fly?

Answer 1

d

Step-by-step explanation:

Answer 2

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

`(V_1)/(C_1) =(V_2)/(C_2) = 0.5`

R = 287 J/kg K

`C_1 = √(\gamma RT_1) = √(1.4* 287* 218) = 295 m/s`

`C_2 = √(\gamma RT_2) = √(1.4* 287* 273) = 331 m/s`

`V_2 = (V_1)/(C_1)*C_2`

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 × 147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁ )

= 165.49 - 147.5

= 17.5 m/s