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)

Final Answer:

x =  9413 m

Step-by-step explanation:

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

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