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Chapter 4 Two-Dimensional Kinematics

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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 – PowerPoint PPT presentation

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Title: Chapter 4 Two-Dimensional Kinematics


1
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
2
4-1 Motion in Two Dimensions
If velocity is constant, motion is along a
straight line
3
4-1 Motion in Two Dimensions
Motion in the x- and y-directions should be
solved separately
4
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

5
4-2 Projectile Motion Basic Equations
The acceleration is independent of the direction
of the velocity
6
4-2 Projectile Motion Basic Equations
These, then, are the basic equations of
projectile motion
7
4-3 Zero Launch Angle
Launch angle direction of initial velocity with
respect to horizontal
8
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
9
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
10
4-3 Zero Launch Angle Example
now that we have an expression for time, t we can
find the range x
11
h 1 m v0X 2 m/s ? 0 x ?
vf
We could also find the final velocity, vf
12
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 !
13
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
14
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
15
4-4 General Launch Angle
In general, v0x v0 cos ? and v0y v0 sin
? This gives the equations of motion
16
4-4 General Launch Angle
Snapshots of a trajectory red dots are at t 1
s, t 2 s, and t 3 s
17
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
18
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
19
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!!!
20
(No Transcript)
21
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.
22
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!
23
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.
24
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!
25
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
26
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.
27
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
28
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?
29
v0??
350
3.3 km
9.4 km
a ½ g b v0 sin ?0 c h
this reduces to
30
v0??
350
3.3 km
9.4 km
now we can substitute this expression for time
into our range equation!
31
v0??
350
3.3 km
9.4 km
Using a lot of algebra and tricks this equation
becomes
32
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
33
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.
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