The aircraft flies at an altitude of 12600 m above the ground at a speed of 532 km/h. At what horizontal distance from point B should be release any body from the aircraft body to fall into point B? (g = 9.81 m/s2)

Correct answer:

x =  7490 m

Step-by-step explanation:

v=532 km/h m/s=532:3.6  m/s=147.77778 m/s g=9.81 m/s2 h=12600 m  h = 21 g t2  t=g2 h=9.812 1260050.6834 s  x=v t=147.7778 50.6834=7490 m

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