Chapter 4 Motion in Two Dimensions EXAMPLES

Example 4.1 Driving off a cliff. yi = 0 at top, y is positive upward.

Also vyi = 0 How fast must the motorcycle

leave the cliff to land at x = 90 m, y = – 50 m

Unknown: vxi = ? Formulas: vy = gt x = vxit y = ½gt2

Time to Bottom:

vxi = x/t = 90.0m/3.19s vxi = 28.2 m/s

2

2 502 3.199.8 /

myt sg m s

Example 4.2 Kicked football

Given: θi = 37º, vi = 20 m/s vxi = vicosθi = 16 m/s & vyi= visinθi = 12 m/s

Find a. Max height (h) ? b. Time when hits ground?

c. Total distance traveled in the x direction (R) ? d. Velocity at top? e. Acceleration at top?

vyi

vxi

vi

Example 4.2 cont.

Using Eq: 4.132 2 2 2( sin ) 12 144 /( ) 7.352 22 2 9 8 19.6 /

( ) At 0, 0. At the end when the ball touches the ground: 0 Using Eq. 4.12:

sin

v ( m/s) m si ia h mg ( . m/s ) m sb t y

y f

y (v θi i

2 2 21 0 (12 / ) 4.9( / ) Solving for t:212 / 0 (of course!) and 2.4524.9 /

( ) ( cos ) 16 2 45 39.2

( ) At this point: 0 16 /

( )

)t gt m s t m s t

m st t sm s

c R v t v θ t ( m/s)( . s) mxi i i

d v v v m sy x xi

e

The acceleration vector is the same at the highest2 point as it is throughout the flight, which is: 9 80

. m/s

Example 4.3 Where Does The Apple Land?

A child sits in a wagon, moving to the right (x-direction) at constant velocity vox. She throws an apple straight up (from her viewpoint) with initial velocity voy while she continues to travel forward at vox Neglect air resistance.

Will the apple land behind the wagon,

in front of the wagon, or in the wagon?

Example 4.3 Cont.

The apple will stay above the girl the entire trip and will land in the wagon.

The reason is: To a person in the ground reference frame (b) the apple will be exactly a projectile in motion (neglecting air resistance). To the girl it is an object in free fall.

And the Vertical motion of a projectile and free fall are the same.

Example 4.4 Wrong Strategy (Similar to Example 4.3 Text Book)

“Shooting the Monkey”!! A boy on a small hill aims his

water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away. At the instant the water balloon is released, the second boy lets go an fall from the tree, hopping to avoid being hit.

Show that he made the wrong move (He hadn’t studied Physics yet!!)

Example 4.4 Cont.

“Shooting the Monkey”!! Both the water balloon and the boy in the tree start falling at

the same time, and in a time t they each fall the same vertical distance y = ½gt2

In the same time it takes the water balloon to travel the horizontal distance d, the balloon will have the same y position as the falling boy.

Splash!!! If the boy had stayed in the tree, he would have avoided the humiliation

Example 4.5 That’s Quite an Arm Non-Symmetric Projectile

Motion Example 4.4 (text book), page 84 Follow the general rules for projectile

motion Break the y-direction into parts

up and down or symmetrical back to initial height and

then the rest of the height May be non-symmetric in other ways

Example 4.5 Cont. Given: θi = 30º, vi = 20 m/s (A) vxi= vicosθi = 17.3 m/s and vyi= visinθi = 10.0 m/s

At t = 0 : xi = 0 yi = 0 Find: t = ? (time at which the stone hits the ground) with yf = –

45.0m Using: yf = vyit – ½gt2

– 45.0m = (10.0m/s)t – (4.90m/s2)t2

Solving for t using General Quadratic Formula: t = 4.22 s (B) vxi= vxf = 17.3 m/s and vyf = vyi – gt

vyf = 10.0m/s – (9.80m/s2)(4.22s) vyf = ̶ 31.4m/s

smsmv

vvvvvv

f

yfxffyfxff

/9.35/)4.31()3.17( 22

22222

Example 4.6 The End of the Ski Jump Example 4.5 (text book), page 85 Given the figure of the ski jumper, find

the distance d traveled along the incline.1. Coordinates x and y at the end:

2. From the figure:

)1(25ttvx xif

)2(9.4 2221 tgty f

)3(35cos dx f

)4(35sin dy f

Example 4.6 Cont.3. Equating (1) = (3) and (2) = (4)

4. Dividing (6) by (5):

5. Substitution of (7) in (5) and solving for d:

6. Substitution of d into (3) and (4), gives the coordinates:

)7(sec57.325

9.435tan tt

)5(2535cos td )6(9.435sin 2td

mdd 109)57.3(2535cos

mx f 3.8935cos)109( my f 5.6235sin)109(

Example 4.7 The Centripetal Acceleration of the Earth Calculate ac of the Earth, assuming it moves in a circular

orbit around the Sun.

Note that ac << g

2

2 2 2 2

2 2

2 11 23 2

2 7

24 4

4 1.496 10 1 5.93 10 /3.156 101

c

c

rv r rTar r rT T

x m yra x m sx syr

Material from the book to Study!!! Objective Questions: 1-4-6 Conceptual Questions: 5-6-7 Problems: 2-4-6-10-11-21-27

Material for the Midterm