# Lesson 2 - y = v iy t 1/2a y t 2 1.7 m = v iy(0.589 s...

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Lesson 2 5) v ax / v a = cos31 v ax = 3.8 m/s v ay / v a = sin31 v ay = -2.3 m/s v bx / v b = cos25 v bx = 7.1 m/s v by / v b = sin32 v by = 3.3 m/s a x = v bx – v ax / t a x = 7.1 m/s - 3.8 m/s / 8.5s a x = 0.4 m/s 2 a y = v bx – v by / t a y = 3.3 m/s + 2.3 m/s / 8.5s a y = 0.7 m/s 2 a 2 = a x 2 + a y 2 a 2 = 0.16 + 0.49 a = +/- 0.8 m/s 2 tan = a y / a x tan = 0.7 / 0.4 tan = 1.75 tan = 60 degrees The bird’s average acceleration is 0.8 m/s 2 60 degrees N of E. 6)a) d x = v x x t t = 9 m / 18 m/s t= 0.5 s b) d y = v iy t + 1/2a y t 2

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d y = ½(9.8 m/s)(0.5) 2 d y   = 1.2 m The snowball hits the tree 1.2 m above the ground. c) v fy = ( d y ) / ( t + ½ a y t 2 ) v fy = (1.2 m) / (0.5 s + ½ (9.8 m/s)(0.25 s) v fy = 0.7 m/s The snowball’s velocity is 0.7 m/s. 7) d y = v fy t - 1/2a y t 2 1.7 m = (0) - ½(-9.8 m/s 2 ) t 2 t 2 = 0.347 s 2 t = 0.589 s d
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Unformatted text preview: y = v iy t + 1/2a y t 2 1.7 m = v iy (0.589 s) + ½(-9.8 m/s 2 )(0.589 s) 2 v iy = 4.6 m/s The minimum vertical velocity of the salmon is 4.6 m/s. 8) Let S represent the swimmer, W the water and G for the ground. ( s v g ) 2 = ( s v w ) 2 + ( w v g ) 2-2( s v w ) ( w v g )cos ( s v g ) 2 = (4 m/s) 2 + (2 m/s) 2-2(4 m/s) (2 m/s)cos65 ( s v g ) 2 = (20 – ( - 6.76)) m 2 / s 2 ( s v g ) 2 = 26.76 m 2 / s 2 ( s v g ) = 5.2 m / s > = cos-1 [(a 2 + b 2 – c 2 ) / (2ab)] > = cos-1 [(16 + 27.04 – 4) / (41.6)] > = cos-1 (0.94) > = 20 The swimmer’s velocity with respect to the crowd is 5.2 m/s [N 20 E]...
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