Page 1 – Cliff Problem

V = 9.21 m/st = 2.17 s Fill in your H/V table of XViVfat 1.How far out does she land? 2.How high is the cliff? 3.What is the velocity of impact in VC Notation? 4.What is the speed of impact? 5.What is the velocity of impact? (in AM Notation)

V = 9.21 m/s t = 2.17 s Fill in your H/V table of XViVfat 1.How far out does she land? 2.How high is the cliff? 3.What is the velocity of impact in VC Notation? 4.What is the speed of impact? 5.What is the velocity of impact? (in AM Notation) HV X?? Vi9.21 m/s0 (cliff) Vf9.21 m/s a0-9.8 m/s/s t2.17 s2.17 s Horiz: X = (9.21)(2.17) = m ≈ 20. m Vert: Vf = Vi + at = 0 + (-9.8)(2.17) = m/s ≈ 21.3 m/s X = V i t + 1/2at 2 = 0 + 1/3(-9.8)(2.17) 2 = m So she lands about 20. m out, the cliff is 23 m tall, and her velocity of impact in VC notation is: 9.21 m/s x m/s y, the speed is the hypotenuse of that √( ) = m/s so her speed is about 23.2 m/s In angle magnitude notation her velocity looks like: The magnitude is the speed we calculated, and the angle indicated is θ = tan -1 (21.266/9.21) = 66.6 o

Page 2 – Arc Problem

V = 29.9 m/s Find vector components Fill in your H/V table of X Vi Vf a t 1.Find the hang time 2.Find the range 3.Find Speed at highest point. 4.Greatest Height the ball reaches angle = 25.0 o

V = 29.9 m/s Find vector components Fill in your H/V table of X Vi Vf a t 1.Find the hang time 2.Find the range 3.Find Speed at highest point. 4.Greatest Height the ball reaches First, resolve the vector into components: Vx = (29.9 m/s)cos(25 o ) = m/s, Vy = (29.9 m/s)sin(25 o ) = m/s – these are your initial velocities. Now we aet up the H/V table HV X ?0 (level ground) Vi27.10 m/s12.64 m/s Vf27.10 m/s m/s (level ground) a0-9.8 m/s/s t?? Vert: Find t using Vi = Vf + at, = (-9.8)t, t = s, which is the hang time Horiz: Find X using X = Vi t = (27.10)(2.579 s) = 69.9 m which is the range At the highest point Vert: Vf = 0 (top) So the speed is purely horizontal = 27.1 m/s in this case, and the greatest height is Vf 2 = Vi 2 + 2aX, 02 = (12.64) 2 + 2(-9.8)X, X = m ≈ 8.15 m angle = 25.0 o

Page 3 - Vectors

Vectors - Finding Components - step by step TOC Step 1: Find the Trig angle – ACW from x axis 0 o Or 360 o 90 o 180 o 270 o 27 o 51 o 15 o 17 o This is the trig angle  T = 270 – 15 = 255 o  T = = 321 o  T = 360 – 17 = 343 o OR  T = -17 o

Whiteboards: Getting the trig angle 11 | 2 | 3 | 4 | 5 | 623 TOC

58 o W What’s the Trigonometric angle? 32 o 90 – 32 = 58 o

218 o W What’s the Trigonometric angle? 38 o = 218 o

257 o W What’s the Trigonometric angle? 13 o = 257 o

158 o W What’s the Trigonometric angle? 22 o = 158 o

116 o W What’s the Trigonometric angle? 26 o = 116 o

318 o W What’s the Trigonometric angle? 42 o = 318 o

Vectors - Try this yourself TOC Get out your calculator type: sin 90 1???? If not MODE Cursor arrows to “Degree” Try again (sin 90)

Vectors - Finding Components - step by step TOC Step 2: Figure the sides using Cos and Sin: x = mag Cos(  ) y = mag Sin(  ) (iff  = trig angle) x = (12 m)Cos ( 153 o ) = m  = 180 o – 27 o = 153 o y = (12 m)Sin ( 153 o ) = m 12 m 27 o

Vectors - Finding Components - step by step TOC Step 3: Write it in Vector Component notation: Vector = -11 m x m y (With sig figs) Reality Check: (+ and -), relative size 12 m 27 o

Vectors - Try this yourself TOC 23.0 m/s 31.0 o 1.Draw the Components 2.Figure the components with sin and cos 3.Write the answer in VC Notation m/s x m/s y  = = 121 o 23cos(121) x + 23sin(121) y x y

Whiteboards: AM to VC 11 | 2 | 3 | 4 | TOC

22.8 km x km y W 32.2 o 27.0 km  = 32.2 o 27.0cos(32.2) x sin(32.2) y km x km y

37 m x m y W 42 m 27 o  = 360 – 27 = 333 o 42cos(333) x + 42sin(333) y m x m y

-4.9 ft x ft y W 5.0 feet 77 o  = = 167 o 5cos(167) x + 5sin(167) y ft x ft y

68 N x N y W N 38 o  = = 308 o 110.0cos(308) x sin(308) y N x N y

-4.3 m/s x m/s y W 5.0 m/s 30.0 o  = = o 5.0cos(210.0) x + 5.0sin(210) y m/s x m/s y

Vectors - VC + VC - step by step TOC Given these Vectors: A: 2.3 m x m y B: 7.4 m x m y A+B = 9.7 m x m y Any Questions?????? (This is why VC vectors are our friends)

Whiteboards: VC + VC 11 | 2 | 3 | 4 | TOC

2.6 m x m y W A = 4.5 m x m y B = -1.2 m x m y C = -1.9 m x m y Find A + C

-3.1 m x +.2 m y W A = 4.5 m x m y B = -1.2 m x m y C = -1.9 m x m y Find C + B

Vectors - VC to AM - step by step TOC Given this VC Vector: 5.1 m x m y 2. Find the angle: 1. Draw the vector: 5.1 m x -1.7 m y Tan  = opp/adj adj opp   = tan -1 (1.7/5.1) = o = 18 o (s.f.) 3. Find the Magnitude: opp 2 + adj 2 = hyp 2 hyp =  ( ) = = 5.4 m

Whiteboards: VC to AM 11 | 2 | 3 | 4 | TOC

3.9 m, 30. o W Draw this vector, and find its magnitude and the angle it forms with the x-axis: (label your angle) 3.4 m x m y 3.4 m 2.0 m 3.9 m 30. o

27 m/s, 56 o W Draw this vector, and find its magnitude and the angle it forms with the y-axis: (label your angle) -22 m/s x + 15 m/s y -22 m/s 15 m/s 27 m/s 56 o

17.5 N, 31.0 o W Draw this vector, and find its magnitude and the angle it forms with the y-axis: (label your angle) 9.00 N x N y 9.00 N N 17.5 N 31 o

8.19, 59.8 o W Add these two angle magnitude vectors, and express their sum as an angle magnitude vector, finding the angle it forms with the x-axis 48.0 o 7.00 A B o 1. AM to VC AM to VC 2. VC + VC 3. AM to VC A = x y B = x y A + B = x y Mag = √( ) = 8.19 θ = Tan -1 (7.073/4.119) = 59.8 o

Page 4 – Boat Crossing River

Current: 1.2 m/s Find: Time to cross Where it lands Speed as seen from above Angle to go straight across? 120 m Boat 3.5 m/s

Current: 1.2 m/s Find: Time to cross Where it lands Speed as seen from above Angle to go straight across? ACDS X120 m? V3.5 m/s1.2 m/s t?? AC: Find time using V = X/t, 3.5 = 120/t, t = s which is true on the DS side too DS: Find X using V = X/t, 1.2 = X/34.29 s, X = m Speed is the hypotenuse of the velocity vector: = √( ) = 3.7 m/s 120 m Boat 3.5 m/s 3.5 m/s 1.2 m/s θ = Sin-1(1.2/3.5) = 20. o