Bomber

The aircraft flies at an altitude of 4100 m above the ground at speed 777 km/h. At what horizontal distance from the point B should be release any body from the aircraft body to fall into point B? (g = 9.81 m/s²)

Correct answer:

x =  6240 m

Step-by-step explanation:

$$\begin{gathered} v=777km\mathrm{/}h=215.833m\mathrm{/}s \\ h={\displaystyle \frac{1}{2}}g{t}^2 \\ t=\sqrt{{\displaystyle \frac{2h}{g}}} \\ t=\sqrt{{\displaystyle \frac{2\cdot 4100}{9.81}}}=28.91s \\ x=vt=v\sqrt{{\displaystyle \frac{2h}{g}}} \\ x=777\mathrm{/}3.6\cdot \sqrt{{\displaystyle \frac{2\cdot 4100}{9.81}}} \\ x=6240\text{ m} \end{gathered}$$

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