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Question 5, chap 3, sect . part 1 of 2 9 points An artillery shell is fired at an angle of 27.4° above the horizontal ground with an initial speed of 1700 m/s. The acceleration of gravity is 9.8 m/s Find the total time of flight of the shell, neglecting air resistance Answer in units of min

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Answer #1

5] Time of flight for a projectile is given by:

T=\frac{2usin\theta}{g}

T=\frac{2(1700)sin27.4\degree}{9.8} = 159.66s = 2.66 minutes

Horizontal range is:

R = u xT = 1700cos(27.4) x 159.66 = 240974.36 m = 240.974 km.

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