Bomber

The aircraft flies at an altitude of 14700 m above the ground at a speed of 619 km/h. At what horizontal distance from point B should be released any body from the aircraft body fall into point B? (g = 9.81 m/s²)

Correct answer:

x =  9413 m

Step-by-step explanation:

$v = 619 \ \text{km/h} \rightarrow \ \text{m/s} = 619 : 3.6 \ \ \text{m/s} = 171.94444 \ \text{m/s} \ \\ g = 9.81 \ \text{m/s}^2 \ \\ h = 14700 \ \text{m} \ \\ \ \\ h\ = \ \dfrac{ 1 }{ 2 } \ g\ t^2 \ \\ \ \\ t = \sqrt{ \dfrac{ 2 \cdot \ h }{ g } } = \sqrt{ \dfrac{ 2 \cdot \ 14700 }{ 9.81 } } \doteq 54.7443 \ \text{s} \ \\ \ \\ x = v \cdot \ t = 171.9444 \cdot \ 54.7443 = 9413 \ \text{m}$

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