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.

Page 1 – Cliff Problem

V = 9.21 m/s t = 2.17 s

Fill in your H/V table of XViVfat1. 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 sFill in your H/V table of XViVfat1. 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)

H VX ? ?Vi 9.21 m/s 0 (cliff)Vf 9.21 m/sa 0 -9.8 m/s/st 2.17 s 2.17 s

Horiz:X = (9.21)(2.17) = 19.9857 m ≈ 20. m

Vert:Vf = Vi + at = 0 + (-9.8)(2.17) = -21.266 m/s ≈ 21.3 m/sX = Vi t + 1/2at2 = 0 + 1/3(-9.8)(2.17)2 = -23.07361 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 + -21.3 m/s y, the speed is the hypotenuse of that √(9.212+21.2662) = 23.1747 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 angleindicated is θ = tan-1(21.266/9.21) = 66.6o

9.21

21.266

Page 2 – Arc Problem

V = 29.9 m/s

Find vector componentsFill in your H/V table of X Vi Vf a t1. Find the hang time2. Find the range3. Find Speed at highest point.4. Greatest Height the ball reaches

angle = 25.0o

V = 29.9 m/s

Find vector componentsFill in your H/V table of X Vi Vf a t1. Find the hang time2. Find the range3. Find Speed at highest point.4. Greatest Height the ball reaches

First, resolve the vector into components:Vx = (29.9 m/s)cos(25o) = 27.10 m/s, Vy = (29.9 m/s)sin(25o) = 12.64 m/s – these are your initial velocities. Now

we aet up the H/V table

H VX ? 0 (level ground)Vi 27.10 m/s 12.64 m/sVf 27.10 m/s -12.64 m/s (level ground)a 0 -9.8 m/s/st ? ?

Vert:Find t using Vi = Vf + at, -12.64 = 12.64 + (-9.8)t, t = 2.579 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 Vf2 = Vi2 + 2aX, 02 = (12.64)2 + 2(-9.8)X, X = 8.147 m ≈ 8.15 m

angle = 25.0o

Page 3 - Vectors

Vectors - Finding Components - step by step

TOC

Step 1: Find the Trig angle – ACW from x axis

0o Or 360o

90o

180o

270o

27o

51o

15o

17o

This is the trig angle

T = 270 – 15 = 255o T = 270 + 51 = 321o

T = 360 – 17 = 343o

OR

T = -17o

Whiteboards: Getting the trig angle

1 | 2 | 3 | 4 | 5 | 6

TOC

58o W

What’s the Trigonometric angle?

32o

90 – 32 = 58o

218o W

What’s the Trigonometric angle?

38o

180 + 38 = 218o

257o W

What’s the Trigonometric angle?

13o

270 - 13 = 257o

158o W

What’s the Trigonometric angle?

22o

180 - 22 = 158o

116o W

What’s the Trigonometric angle?

26o

90 + 26 = 116o

318o W

What’s the Trigonometric angle?

42o

360 - 42 = 318o

Vectors - Try this yourself

TOC

Get out your calculatortype:sin 90 <ENTER>1????If not <2nd?> MODECursor arrows to “Degree”<ENTER> <CLEAR>Try again (sin 90)

Vectors - Finding Components - step by step

TOC

Step 2: Figure the sidesusing Cos and Sin:x = mag Cos()y = mag Sin()(iff = trig angle)

x = (12 m)Cos(153o) = -10.692 m

= 180o – 27o = 153o

y = (12 m)Sin(153o) = +5.448 m

12 m

27o

Vectors - Finding Components - step by step

TOC

Step 3: Write it in Vector Component notation:

Vector = -11 m x + 5.4 m y (With sig figs)Reality Check:

(+ and -), relative size

12 m

27o

Vectors - Try this yourself

TOC

23.0 m/s31.0o

1. Draw the Components2. Figure the components with sin and cos3. Write the answer in VC Notation

-11.8 m/s x + 19.7 m/s y

= 90 + 31 = 121o

23cos(121) x + 23sin(121) y-11.846 x + 19.715 y

Whiteboards: AM to VC

1 | 2 | 3 | 4 | 5

TOC

22.8 km x + 14.4 km y W

32.2o

27.0 km

= 32.2o

27.0cos(32.2) x + 27.0sin(32.2) y22.84721549 14.3876594522.8 km x + 14.4 km y

37 m x + -19 m y W

42 m

27o

= 360 – 27 = 333o

42cos(333) x + 42sin(333) y37.42227402 -19.0676009937 m x + -19 m y

-4.9 ft x + 1.1 ft y W

5.0 feet77o

= 90 + 77 = 167o

5cos(167) x + 5sin(167) y-4.871850324 1.124755272-4.9 ft x + 1.1 ft y

68 N x + -87 N y W

110.0 N38o

= 270 + 38 = 308o

110.0cos(308) x + 110.0sin(308) y67.72276229 -86.681182968 N x + -87 N y

-4.3 m/s x + -2.5 m/s y W

5.0 m/s

30.0o

= 180 + 30.0 = 210.0o

5.0cos(210.0) x + 5.0sin(210) y-4.330127019 -2.5-4.3 m/s x + -2.5 m/s y

Vectors - VC + VC - step by step

TOC

Given these Vectors:

A: 2.3 m x + 3.4 m yB: 7.4 m x + 1.2 m yA+B = 9.7 m x + 4.6 m y

Any Questions??????(This is why VC vectors are our friends)

Whiteboards: VC + VC

1 | 2 | 3 | 4 | 5

TOC

2.6 m x + 7.3 m y W

A = 4.5 m x + 3.2 m yB = -1.2 m x + -3.9 m yC = -1.9 m x + 4.1 m y

Find A + C

-3.1 m x + .2 m y W

A = 4.5 m x + 3.2 m yB = -1.2 m x + -3.9 m yC = -1.9 m x + 4.1 m y

Find C + B

Vectors - VC to AM - step by step

TOC

Given this VC Vector: 5.1 m x + -1.7 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) = 18.435o = 18o (s.f.)

3. Find the Magnitude:opp2 + adj2 = hyp2 hyp = (5.12 + 1.72) = 5.376 = 5.4 m

Whiteboards: VC to AM

1 | 2 | 3 | 4 | 5

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 + 2.0 m y

3.4 m

2.0 m3.9 m

30.o

27 m/s, 56o 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/s56o

17.5 N, 31.0o W

Draw this vector, and find its magnitude and the angle it forms with the y-axis: (label your angle) 9.00 N x + -15.0 N y

9.00 N

-15.0 N17.5 N

31o

8.19 , 59.8o 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

48.0o

7.00

A B

9.00

78.0o

1. AM to VC AM to VC2. VC + VC3. AM to VC

A = 4.684 x + 5.202 yB = -8.803 x + 1.871 yA + B = -4.119 x 7.073 y

Mag = √(4.1192+7.0732) = 8.19

θ = Tan-1(7.073/4.119) = 59.8o

Page 4 – Boat Crossing River

Current: 1.2 m/sFind:

•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?

AC DSX 120 m ?V 3.5 m/s 1.2 m/st ? ?

AC: Find time using V = X/t, 3.5 = 120/t, t = 34.29 s which is true on the DS side too

DS:Find X using V = X/t, 1.2 = X/34.29 s, X = 41.14 m

Speed is the hypotenuse of the velocity vector:= √(3.52+1.22) = 3.7 m/s

120 m

Boat 3.5 m/s

3.5 m/s1.2 m/s

θ = Sin-1(1.2/3.5) = 20. o