1 Physics

1. Vi x = 90 cos 53 = 90 x 0.602 = 54.18 m/s Vi y = 90 blunder 53 = 90 x 0.799 = 71.91 m/s A)Y = ( vi y x t ) + ½ at^2 0 = 71.91 x t (4.9 m/s^2) x t^2 t = 71.91/4.9 = 14.68 = 14.7s B)V = vi + at 0 = Vi y - 9.8 t in that location for t = 71.91/9.8 = 7.338 = 7.34s Y = Vi y x t 4.9 x t^2 = 71.91 x 7.338 ( 4.9x(7.338)^2) = 264m C)X = Vix t = 54.18x14.68 = 795m D)Vfy = Viy 9.8t = 71.91 98.8 x 14.68 = -71.95 m/s Vfx = Vix = 54.18 m/s Vf = Vfy ^2 + vfx^2 = 90 now to husking an angel of it Tan theta = vfy / vfx = -71.91/54.18 = -1.328 Theta = tan^-1 -1.328 = -53 Vf = 90.0 m/s @ -57 2. Vix = 30.0 cos 30 = 30x0.866 = 25.98 m/s Viy = 30.0 pit 30 = 30 x 0.5 = 15.0 m/s a)Y = 15 x 1.531 4.9 x 1.531^2 = 11.48 m There for numerate hight would be 200m + 11.48m = 211.48m = 211m b)Vfy = Viy 9.8 t 0 = 15 9.8t t = 15/9.8 = 1.531 = 1.53s c)X = Vix t = 25.98 x 8.1 = 210m d)Vfy ^2 = Viy^2 + 2 x -9.8 Yf Vfy ^2 = 15^2 + 2 x -9.8 x -200 = 4145 Vfy = 64.38 m/s e)Vfx = Vix =25.98 Vf = (Vfx^2 + Vfy^2)^ ½ = 69.4 at one clock time to go steady angle Tan theta = vfy/vfx = -64.28/25.98 = -2.474 Theta = tan^-1 -2.474 = -68 There for vf = 69.4m/s @ -68 3. a) Y = Viyt ½ 9.8 t^2 -98 = 0-4.9t^2 There for t = (98/4.9)^ ½ = 4.472 = 4.47s b) Xf = Vxi x t = 40 x 4.472 =4.47s c) Vyf = vyi 9.8t = 0-9.8x4.472 = -43.

83 m/2 now Vf^2 = Vyf^2 + Vxf^2 =43.83^2 + 40.0^2 = 1921 + 1600 = 3521 there for vf = 59.3 m/2 now to find angle tan theta = vyf/vxf = -43.83/40.0 = -1.096 there for theta = tan^-1 -1.096 = -48 there for vf = 59.3m/s @ -48 d) x = 179m, y = 98m D = (x^2 + y^2)^ ½ = (1! 79^2 + 98^2)^ ½ = 204 Tan theta = 98/179 = 0.547 There for theta = tan^-1 0.547 = 29 D = 204m @ -29 4) Vix = 35.0 cos 37 = 35 x 0.799 = 27.97 m/s Viy = 35.0 sin 37 = 35 x 0.602 = 21.07 m/s a)y = 21.07 x 2.15 4.9x2.15^2 = 22.65 there for highest total hight would be 300+22.65 = 322.65 = 323 b)Vfy = Viy 9.8 t 0 = 21.07 9.8 t t = 21.07/9.8 = 2.15s...If you requisite to get a full essay, give it on our website: OrderCustomPaper.com

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