An artillery shell is fired at an angle of 62.7 above the horizontal ground with an initial speed of 1520 m/s. The acceleration of gravity is 9.8 m/s2 . Find the total time of flight of the shell, neglecting air resistance. Answer in units of min

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vertical is 1520sin62.7 = 1350.69819 horizontal is 1520cos62.7 = 697.14732 s=ut +1/2 at^2 therefore 1350.698t +4.9t^2 t=277.77 seconds... and then divide 697.1432/277.77 = part b Can someone tell me if this is the correct answer, or if not what I am doing wrong.

isnt this a math question?

nope... university physics

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