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Unformatted text preview: h ic e8 e k# i b iC a aha 3 Problem 12: Let x point to the right and y point up. We have Horizontal: Vertical: x = 1 . 75 m y = 9 . 00 m v x = v = ? v y = 0 m / s v fx = ? v fy = ? a x = 0 m / s 2 a y = g = 9 . 8 m / s 2 t = 0 t = 0 We have enough information in the vertical to find t : y = arrownortheast v y t + 1 2 a y ( t ) 2 t = radicalBigg 2 y a y = radicalBigg 2( 9 . 00 m) ( 9 . 8 m / s 2 ) = 1 . 36 s . Now we use the horizontal information to find v x : x = v x t + 1 2 arrownortheast a x ( t ) 2 v = v x = x t = 1 . 75 m 1 . 36 s = 1 . 29 m / s . 3 Problem 13 ( a ): This is very similar to Problem 3.12 above. We let x point to the right and y point up. We have Horizontal: Vertical: x = 61 . 0 m y = (1 . 8 m) (21 . 3 m) = 19 . 5 m v x = v = ? v y = 0 m / s v fx = ? v fy = ? a x = 0 m / s 2 a y = g = 9 . 8 m / s 2 t = 0 t = 0 We use the vertical information to find t : y = arrownortheast v y t + 1 2 a y ( t ) 2 t = radicalBigg 2 y a y = radicalBigg 2( 9 . 00 m) ( 9 . 8 m / s 2 ) = 1 . 99 s . Now we use the horizontal information to find v x : x = v x t + 1 2 arrownortheast a x ( t ) 2 v = v x = x t = 61 . 0 m 1 . 99 s = 30 . 7 m / s . ( b ): Now we find the final velocities. v fx = v x + a x t = v x = 30 . 7 m / s v fy = arrownortheast v y + a y t = ( 9 . 8 m / s 2 )(1 . 99 s) = 19 . 5 m / s . Thus the final speed is v f = radicalBig v 2 fx + v 2 fy = radicalbig (30 . 7 m / s) 2 + ( 19 . 5 m / s) 2 = 36 . 4 m / s . 3 Problem 27: The suitcase starts with an initial velocity of v = 90 . 9 m / s at an angle = 23 . above the ground, and falls a distance 114 m down. If we let the yaxis point upwards, then we have Horizontal: Vertical: x = ? y = 114 m v x = v cos = 82 . 8 m / s v y = 35 . 2 m / s v fx = ? v fy = ? a x = 0 m / s 2 a y = g = 9 . 8 m / s 2 t = 0 t = 0 1 We use the vertical information to solve for the time: y = v y t + 1 2 a y ( t ) 2 = v sin t 1 2 g ( t ) 2 1 2 g ( t ) 2 v sin t + y = 0 . Using the quadratic formula, we get t = v sin radicalBig v 2 sin 2  2 g y g = braceleftBig 9 . 59 s 2 . 42 s . We want the positive solution. In this amount of time the suitcase travels horizontally a distance x = v x t + 1 2 arrownortheast a x ( t ) 2 = v cos t = 794 m ....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
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