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Chapter 4 Two-Dimensional Kinematics

- Learning Objectives
- Motion in Two Dimensions
- Projectile Motion Basic Equations
- Zero Launch Angle
- General Launch Angle
- Projectile Motion Key Characteristics

PowerPoint presentations are compiled from Walker

3rd Edition Instructor CD-ROM and Dr. Daniel

Bullocks own resources

4-1 Motion in Two Dimensions

If velocity is constant, motion is along a

straight line

4-1 Motion in Two Dimensions

Motion in the x- and y-directions should be

solved separately

4-2 Projectile Motion Basic Equations

- Assumptions
- ignore air resistance
- g 9.81 m/s2, downward
- ignore Earths 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

now that we have an expression for time, t we can

find the range x

h 1 m v0X 2 m/s ? 0 x ?

vf

We could also find the final velocity, vf

Motion in more than one dimension. http//www.phys

icsclassroom.com/mmedia/vectors/mzi.html

straight line (line of site)

the monkey begins to fall ad the precise moment

when the ball leaves the barrel of the gun The

ball and the monkey arrive at the point marked by

the red dot at the same time

path of bullet

path of monkey

Two types of motion occurring !

Monkey Object under constant acceleration in

1 dimension

y

Bullet The motion of the bullet is a

combination of motion with a constant velocity

(along the x-axis) and motion with constant

acceleration (along the y-axis)

x

To start analyzing this problem we must find the

x, and y components of the velocity

vi

y

vi hyp

viy opp

?

vix adj

x

Let start by looking at an example where the

launch angle is zero

4-4 General Launch Angle

In general, v0x v0 cos ? and v0y v0 sin

? 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

Lets 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

harbor entrance 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?

We need an expression for the time, t

We need an expression for the time, t

since the projectile will

return to the same height that it left from, h

y y0 0 so our equation becomes

from here we can solve for time, t

factor out a t, and it cancels

now substitute this back into

from trigonometry

this is called the Range equation!

DANGER WILL ROBINSON This equation only works if

the projectile returns to the same height that it

was launched!!!

(No Transcript)

Now lets look at an example of projectile motion

where the projectile lands at a different

elevation from its launch height.

A stone is projected at a cliff of height h with

an initial speed of 42 m/s directed at angle ?0

600 above the horizontal. The stone strikes A 5.5

s after launching. Find h, the speed of

the stone just before impacting A, and the

maximum height H reached above the ground.

total time, t 5.5 s

v0 42 m/s

600

Lets start by finding the v and y components of

the initial velocity, v0 (Note this is always a

good place to start!!)

Now we can proceed with finding the height of

the cliff!

total time, t 5.5 s

v0 42 m/s

v0y36.4 m/s

600

v0x21 m/s

Now lets try and find the speed of the rock just

before impact. Remember that this will be the

magnitude of the final velocity vector. And this

vector has both an x and y component.

total time, t 5.5 s

v0 42 m/s

vf

v0y36.4 m/s

52m

600

v0x21 m/s

Because there is no acceleration along the x

axis v0xvfx21 m/s However in the y-direction

there is acceleration so we must find the y

component of the final velocity

This negative sign means the y-component is

downward!

total time, t 5.5 s

v0 42 m/s

vf27.3 m/s

v0y36.4 m/s

52m

600

v0x21 m/s

Now lets find the maximum height, H! To do this

you have to know that the instant the stone is at

its maximum height the y component of the

velocity equals zero, vyMAV H 0. Using this

information we can use

350

3.3 km

9.4 km

During volcano eruptions, chunks of solid rock

can be blasted out of the volcano these

projectiles are called volcanic bombs. At what

initial speed would a bomb have to be ejected, at

angle ?0350 to the horizontal, from the vent at

A in order to fall at the foot of the volcano at

B, at vertical distance h3.3 km and horizontal

distance d9.4 km ?

So we need to do is figure at what the initial

speed of the bombs need to be in order to hit

point B.

v0??

350

3.3 km

9.4 km

We have an equation for range, or horizontal

distance

We have, x 9.4 km, we have ?0350 what we need

is the time, t, and this is where it get tricky!

Lets start by using

v0??

350

3.3 km

9.4 km

Im going to bring the h over and make this a

quadratic equation

a

b

c

What is the solution to a quadratic equation?

v0??

350

3.3 km

9.4 km

a ½ g b v0 sin ?0 c h

this reduces to

v0??

350

3.3 km

9.4 km

now we can substitute this expression for time

into our range equation!

v0??

350

3.3 km

9.4 km

Using a lot of algebra and tricks this equation

becomes

v0255 m/s

t ?

350

3.3 km

9.4 km

Next, I would like to find the time of flight. We

can rearrange this equation to solve for time

v0255 m/s

t 45 s

350

3.3 km

9.4 km

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.