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)

Result

x =  6240 m

Solution:

v=777 km/h=215.833 m/s h=12gt2 t=2hg t=241009.81=28.91 s x=vt=v2hg x=777/3.6241009.81 x=6240 mv = 777 \ km/h = 215.833 \ m/s \ \\ h = \dfrac12 g t^2 \ \\ t = \sqrt{ \dfrac{2h}{g}} \ \\ t = \sqrt{ \dfrac{2\cdot 4100 }{ 9.81}} = 28.91 \ s \ \\ x = v t = v \sqrt{ \dfrac{2h}{g}} \ \\ x = 777 / 3.6 \cdot \sqrt{ \dfrac{2\cdot 4100 }{ 9.81}} \ \\ x = 6240 \ \text{m}



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