182_Physics ProblemsTechnical Physics

182_Physics ProblemsTechnical Physics - 184 P6.66 Circular...

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184 Circular Motion and Other Applications of Newton’s Laws P6.66 Let the x –axis point eastward, the y -axis upward, and the z -axis point southward. (a) The range is Z v g i i = 2 2 sin θ The initial speed of the ball is therefore v gZ i i = = ° = sin . sin . . 2 9 80 285 96 0 53 0 θ a fa f m s The time the ball is in the air is found from y v t a t iy y = + 1 2 2 as 0 53 0 48 0 4 90 2 = ° . sin . . m s m s 2 b ga f e j t t giving t = 8 04 . s . (b) v R ix e i = = × ° = 2 86 400 2 6 37 10 35 0 86 400 379 6 π φ π cos . cos . s m s m s e j (c) 360° of latitude corresponds to a distance of 2 π R e , so 285 m is a change in latitude of φ π π = F H G I K J ° = × F H G G I K J J ° = × S R e 2 360 285 6 37 10 360 2 56 10 6 3 a f e j a f m 2 m degrees . . The final latitude is then φ φ φ f i = = °− °= ° 35 0 0 002 56 34 997 4 . . . . The cup is moving eastward at a speed v R fx e f = 2 86 400 π φ cos
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