3 Chapter 3 Problem Assignment

3 Chapter 3 Problem Assignment - Chapter 3 Problem...

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Unformatted text preview: Chapter 3 Problem Assignment 1. Work: Sin 45=x15=10.6 m/s Vo = 10.6 m/s a = -9.8 m/s2 x=xo+Vot+12at2 0m = 0m + (10.6m/s)t + ½ (-9.8m/s2)t2 0 = 10.6m/s t -4.9t2 4.9m/s2t2=10.6 m/s t 4.9m/s2 t = 10.6 t = 2.16 s t = 2.16s/2 t = 1.08s Answer: 1.08 s 2. Vo = 100 m/s t=2s x = vt x = (100 m/s)2s x = 200m Answer: 200 m 3. a2+b2=c2 1002+1002=c2 10,000 + 10,000 = c2 √20,000 = c c = 141.4213562 N a = net force/mass a = 141.42N/50kg a = 2.8284 N/kg2 Answer: 2.82 N/kg2 4. Magnitude 5. a2+b2=c2 52+ 42=c2 25 + 16 = c2 √41 = c c = 6.40 m Answer: 6.40 m 6. x=xo+Vot+12at2 0m = 0m + (19 m/s)t + ½ (-9.8m/s2)t2 0m = 19m/s t – 4.9m/s2 t2 4.9m/s2 t2 = 19m/s t 4.9m/s2 t = 19m/s t = 3.88 s t = 3.88s/2 t = 1.94 s x = 0 +19m/s(1.94s) + ½(-9.8m/s2)(1.94)2 x = 18.4m Answer: 18.4 m 7. Vo = 0 m/s, xo = 25m, x = 0m, a = -9.8m/s2 0m = 25m + 0m/s t + ½(-9.8m/s2)t2 0 = 25 – 4.9t2 4.9t2 = 25 t = 2.26 s x = 0 + 35(2.26) + ½ (0)(2.26)2 x = 35(2.26) x = 79.1 m Answer: 2.26s for the ball to strike the water; 79.1 m to land horizontally from the bridge 8. g = 9.8m/s2, theta = 37*, x = 35 m, and y = 2.84 m 2.84 = 35 tan(37) – 35 x 35g/(2v2cos(37)2 2.84 = 26.25-1225(10) x 252 x 4 x 4v2 23.41 = 1225(10) x 252 x 4 x 4v2 v2=400.01 v = 20 m/s Answer : 20 m/s 9. Cos 45=x10→ Ax=7.07 m, Ay=7.07m 60=x12→Bx=6m a2+b2=c2 62+ b2=122 36+b2=144 b2=108 By = -10.4 m 7.07x + 6.00x = 13.07x 7.07y + -10.4y = -3.33y 13.072±3.332=c2 170.8249+11.6889=c2 181.9138=c2 x===.135 = Answer: 13.5 m 10. Cos 30 = x50 = 43.3 m/s Sin 30 = x50 = 25 m/s x = 0 + 25t + 12-9.8t2 0=25t-4.9t2 4.9t2=25t 4.9t=25 x==.51 ¨ ==.512 ==.255 . F ==+0252 +12-9.8(2.55)2 .55 ˜ ===.6 −3 .753186225 ¨ ===.319 = Answer: 31.9 m 11. ...
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This note was uploaded on 07/11/2011 for the course PHYS 225 taught by Professor Bacon during the Summer '09 term at Ferris State.

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