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

Answer

Assuming

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

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

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$

R = 287 J/kg K

$C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s$

$C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s$

$V_2 = \dfrac{V_1}{C_1}\timesC_2$

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

Answer 2

Answer:

d

Explanation: