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Given: (BJC 10th 11.169) At the bottom of the loop in the vertical plane, a jet has a horizontal veloce shown. In addition is

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SOLUTION: 462 5280 a d y = 45.425 ft/ dt² - lya180oft x= 2400ft re= sehry - - de xdx + ydy ) at at 2 From 7 Tu=52400²+ 180021 PPNI Now tano = y 7 8= tan (oha) - Also d (tano) = at 7 seche x do = x. d y yo dun e dit au do = x. du - y. da – 6 x²sec²o

FROM A = tant/ 190 2400 - tam 36.87 From 6 do - 240000 - 18000 462 2400 ²x 8ec236.87 do = – 0.0924 vals at from 2x0.75) X(0.0

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