Chapter 4 Two-Dimensional Kinematics

PowerPoint presentations are compiled from Walker 3rd Edition Instructor CD-ROM and Dr. Daniel Bullock’s own resources

Learning Objectives

• Motion in Two Dimensions

• Projectile Motion: Basic Equations

• Zero Launch Angle

• General Launch Angle

• Projectile Motion: Key Characteristics

4-1 Motion in Two Dimensions

If velocity is constant, motion is along a straight line:

Motion in the x- and y-directions should be solved separately:

4-1 Motion in Two Dimensions

4-2 Projectile Motion: Basic Equations

Assumptions:

• ignore air resistance

• g = 9.81 m/s2, downward

• ignore Earth’s rotation

If y-axis points upward, acceleration in x-direction is zero and acceleration in y-direction is -9.81 m/s2

4-2 Projectile Motion: Basic Equations

The acceleration is independent of the direction of the velocity:

4-2 Projectile Motion: Basic Equations

These, then, are the basic equations of projectile motion:

4-3 Zero Launch Angle

Launch angle: direction of initial velocity with respect to horizontal

4-3 Zero Launch Angle

In this case, the initial velocity in the y-direction is zero. Here are the equations of motion, with x0 = 0 and y0 = h:

4-3 Zero Launch Angle

Eliminating t and solving for y as a function of x:

This has the form y = a + bx2, which is the equation of a parabola.

The landing point can be found by setting y = 0 and solving for x:

4-3 Zero Launch Angle Example

g

ht

g

httgh

ttimeforsolvetgh

tgtgtvhv

tgtvh

tvx

yy

y

x

222

21

21

21

0

21

22

2

2200

20

0

,

now that we have an expression for time, t we can find the range x

h = 1 mv0X = 2 m/s = 0

x = ?

m

smm

s

m

g

hvx

g

ht

tvx

x

x

9089

122

2

2

2

0

0

..

We could also find the final velocity, vf

vf

??

fy

xfx

fyfxf

fyfxf

vs

mvv

vvv

yvxvv

20

22

s

m

ss

m

gtvv yfy

54

450890 2

0

.

..

s

m

s

m

s

m

vvv fyfxf

84

54222

22

.

.

Motion in more than one dimension.http://www.physicsclassroom.com/mmedia/vectors/mzi.html

straight line(line of site)

path ofbullet

path of monkey

the monkey begins to fall ad theprecise moment when the ballleaves the barrel of the gun

The ball and themonkey arrive atthe point markedby the reddot at thesame time

Two types of motion occurring !

Monkey: Object under constant acceleration in 1 – dimension

Bullet: The motion of the bullet is a combinationof motion with a constant velocity (alongthe x-axis) and motion with constant acceleration (along the y-axis)

y

x

x

y

jvivv iyixi

vi

viy opp

vix adj

coscos

sinsin

iixi

ix

iiyi

iy

vvv

v

hyp

adj

vvv

v

hyp

opp

|v i| h

yp

To start analyzing this problem we must find the x, and y componentsof the velocity:

Let’ start by looking at an example where the launch angle is zero

4-4 General Launch Angle

In general, v0x = v0 cos θ and v0y = v0 sin θ

This gives the equations of motion:

4-4 General Launch Angle

Snapshots of a trajectory; red dots are at t = 1 s, t = 2 s, and t = 3 s

Let’s work an example that is a launch angle different from 0, but it returns to the same height that it is launched.

A pirate ship is 560 m from a fort defending the harborentrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/s

At what angle, from the horizontal must a ball be fired to hit the ship?

tv

tvx x

cos0

0

We need an expression for the time, t

tv

tvx x

cos0

0

We need an expression for the time, t

200 2

1gttvyyh y since the projectile will

return to the same height that it left from,h = y –y0 = 0 so our equation becomes:

20

20 2

121

0 gttvgttv y sin

20 2

10 gttv sin from here we can solve for time, t

factor out a t, and it cancels: sinsin 00 21

21

0 vgtgtv

g

vtvgt

sinsin 0

0

22 now substitute this back into:

tvx cos0

sincossin

coscosg

v

g

vvtvx

200

00

22

from trigonometry: cossinsin 22

220 sing

vx this is called the Range equation!

DANGER WILL ROBINSON: This equation only works if the projectilereturns to the same height that it was launched!!!

8160

82

5608922 2

2

20

20

.

/

/.sinsin

sm

msm

v

gx

g

vx

oo or

or

6327

3125654180654

81602 1

...

.sin

Now let’s look at an example of projectile motion where the projectile lands at a different elevation from it’s launch height.

A stone is projected at a cliff of height h with an initial speedof 42 m/s directed at angle 0 = 600 above the horizontal. Thestone strikes A 5.5 s after launching. Find: h, the speed of thestone just before impacting A, and the maximum height H reachedabove the ground.

v0= 42 m/s

= 600

total time, t = 5.5 s

Let’s start by finding the v and y components of the initial velocity,v0 (Note: this is always a good place to start!!)

s

m

s

mvv

v

v

hyp

adj

s

m

s

mvv

v

v

hyp

opp

xx

yy

216042

4366042

000

0

0

000

0

0

coscoscos

.sinsinsin

s

mv

s

mv xy 21436 00 . Now we can proceed with

finding the height of the cliff!

mmm

ss

ms

s

mgttvh y

5221482200

558921

5543621 2

22

0

..

....

v0= 42 m/s

= 600

total time, t = 5.5 s

s

mv

s

mv xy 21436 00 .

v0y=36.4 m/s

v0x=21 m/s

Now let’s try and find the speed of the rock just before impact.Remember that this will be the magnitude of the final velocityvector. And this vector has both an x and y component.

v0= 42 m/s

= 600

total time, t = 5.5 s

v0y=36.4 m/s

v0x=21 m/s

52m

Because there is no acceleration along the x axis: v0x=vfx=21 m/sHowever in the y-direction there is acceleration so we must findthe y component of the final velocity:

s

ms

s

m

s

mgtvv yfy 5175589436 20 ....

This negativesign means they-componentis downward!

vf

s

m

s

m

s

mvvv fxfyf 32721517

2222 ..

v0= 42 m/s

= 600

total time, t = 5.5 s

v0y=36.4 m/s

v0x=21 m/s

52m

vf=27.3 m/s

Now let’s find the maximum height, H! To do this you have to knowthat the instant the stone is at it’s maximum height the y componentof the velocity equals zero, vyMAV H = 0. Using this information we canuse:

m

smsm

g

vHvgH

gHvgHvv

yy

yyfy

567892

436

22

202

2

2

202

0

20

20

2

..

.

During volcano eruptions, chunks of solid rock can be blasted out of thevolcano; these projectiles are called volcanic bombs. At what initial speedwould a bomb have to be ejected, at angle 0=350 to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at verticaldistance h=3.3 km and horizontal distance d=9.4 km ?

= 350

So we need to do is figure at what the initial speed of thebombs need to be in order to hit point B.

= 9.4 km

= 3.3 km

= 350

= 9.4 km

= 3.3 km

v0=??

We have an equation for range, or horizontal distance:

t

xvtvtvx x

00000

cos

cos

We have, x = 9.4 km, we have 0=350 what we need is the time, t, andthis is where it get tricky! Let’s start by using:

200

200 2

121

gttvhgttvhyy y sin

= 350

= 9.4 km

= 3.3 km

v0=??

200 2

1gttvh sin I’m going to bring the –h over and make

this a quadratic equation

021

002 htvgt sin

a b c

02 cxbxaWhat is the solutionto a quadraticequation?

= 350

= 9.4 km

= 3.3 km

v0=??

a

acbbxcxbxa

24

02

2

021

002 htvgt sin

a = ½ gb= v0 sin 0

c = h

g

hgv

vt

21

2

21

4200

00

sin

sin

this reducesto

= 350

= 9.4 km

= 3.3 km

v0=??

g

ghvv

g

hgv

vt

2

21

2

21

4

022

000

200

00

sinsin

sin

sin

now we can substitute this expression for time into our range equation!

= 350

= 9.4 km

= 3.3 km

v0=??

t

xv

00 cos

g

ghvvt

2022

000

sinsin

Using a lot of algebra and tricks this equation becomes:

s

m

hx

gxv 5255

2 000 .

tancos

= 350

= 9.4 km

= 3.3 km

v0=255 m/s

Next, I would like to find the time of flight. We can rearrange thisequation to solve for time:

s

sm

m

v

x

v

xttvx

xx 45

35255

94000000

0 coscos

t= ?

= 350

= 9.4 km

= 3.3 km

v0=255 m/s

t= 45 s

Finally how would air resistance change our initial velocity?

We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target.