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–/1 pointsSerPSE7 4.P.018. Ask Your Teacher My Notes Question PartSubmissions Used A dive bomber has a...

–/1 pointsSerPSE7 4.P.018.

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A dive bomber has a velocity of 280 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.15 km. Find the angle θ.
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Sol According to given data free body diagram con be drawn as follows 280m 3.15 km 2.15 km 1 2150m - 315om GROUND from figurecos o = x 3150 x = 3150xcos o x = 3150 Xcos (43.030) [x = 2302.63m by equation of motion we have 2 = 1 at txot to 2302.63=0x2ملف فيه - و ه در ه ا -و ۹۰+ او د = = = x9.81xt²+280xsinoxt 2 ) 1 1903 - 55 گردد. 2) 9906 - s *) و ۵ روت و ردود ( ( 1876 / 6 /are get 2150 = 331.42 x (11 tona) + 2301.5x tano 2150 = 331.42 + 331.4 2 tonot 2301.6 tona 331.42 tono +2301.6 tanot - 1818-5

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