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 ii = 2 2 sin θ The initial speed of the ball is therefore v gZ i i == ° = sin . sin . . 2 980 285 96 0 53 0 af ms The time the ball is in the air is found from yv t a t iy y =+ 1 2 2 as 0 53 0 48 0 4 90 2 .s i n .. 2 bg ej tt giving t = 804 . s . (b) v R ix ei ×° = 2 86 400 2 6 37 10 35 0 86 400 379 6 πφ π cos .c o s . s m s (c) 360° of latitude corresponds to a distance of 2 R e , so 285 m is a change in latitude of φ = F H G I K J °= × F H G G I K J J × S R e 2 360 285 637 10 360 2 56 10 6 3 m 2 m degrees . . The final latitude is then φφ fi =− = ° ° = ° 35 0 0 002 56 34 997 4 . . The cup is moving eastward at a speed v R fx ef = 2 86 400 cos s , which is larger than the eastward
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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