Transcript Slide 1

up at an angle q
IV.Projectile fired __________________________
speed vi
with an initial _________________
air resistance
Assume no _________________.
The only force
gravity . This means
acting on the projectile is _________
downward
-9.81 m/s2 ______________
the acceleration is ____________,
-9.81 m/s2
≠0
vtop ______,
atop = ___________
The velocity
is always
tangent to
__________
the path
gravity
vi
max. height
q
the range
v
To solve the problem,
vi must be ____________
resolved
viy =
visinq
into its horizontal (vix)
_______
and vertical (viy)
components
_____________________.
vi
q
vicosq
vix =____________
2 + v 2)
√(v
Where: vi = _______________
is the initial speed,
ix
iy
tan-1(viy/vix)
and q = __________________
is the angle.
2 simultaneous motions:
There are _____
x motion, use: _____________________
dx, vx and ax
For ___
dy, vy and ay
For ___
y motion, use: _____________________
vix =
A. The horizontal motion is determined by ___
vicosq . Because there is _______
no
_______
horizontal force,
uniform
remains constant  _____________
vix __________________
x-motion.
acceleration:
ax =
0
ax
t
velocity:
vfx = vix
vfx
t
displacement:
dx = vixt
dx
t
viy = _______
visinq .
B. Vertical motion is determined by ___
gravity
Because of ____________,
the y motion is like a ball
viy .
straight up
thrown _______________
with an initial speed ____
acceleration: ay = -g
= -9.8 m/s2
velocity:
vfy = viy + ayt
ay
-g
t
viy
= visinq – 9.8t
vfy
dy = viyt + (1/2)ayt2
dy
t
displacement:
= visinqt – (1/2)9.8t2
t
Ex 1: Ms. Rudd is fired out of a cannon at a speed
of 75 m/s and at an angle of 370 to the horizontal.
viy
75 m/s
370
vix
vix = vicosq = (75 m/s)cos370
= 60. m/s
viy = visinq = (75 m/s)sin370
= 45. m/s
To determine how high up she goes and how long
straight
she is in the air, "pretend" she is fired __________
viy = __________
up but with an initial speed = _____
___
45. m/s
Given: viy = 45. m/s
1st Unknown:
dy
ay =-9.8 m/s2
vfy = 0
2nd Unknown:
t
How far up?
vf2 = vi2 + 2ad
02 = 452 + 2(-9.81)d
103 m = d
How long is she in the air?
vf = vi + at
0 = 45+ (-9.8)t
4.6 s = t
0 , this t represents the
Because we chose vfy = ___
rise only . To get the total time
time to _________________
of flight, we must _____________________
. So, the
double this time
9.2
total time t = _______
s. You could get this time
-45 m/s . Then:
directly if you assume vfy = __________
vf = vi + at
-45 = 45 + -9.8 t
9.2 s = t
To determine her range, you must assume her
vix = 60.
m/s.
uniform
x motion is ____________
at vi = ____
_______
Given: vix =
60. m/s
ax =
0
t=
9.2 s
Unknown:
dx
total time is used here!
Notice that the ___________
d = vit + (1/2)at2
= (60.)9.2 s +
=
550 m
0
air resistance
With no ____________________,
only the force of
gravity acts on the object:
___________
vi
gravity
The
trajectory (path)
parabola
is a________________.
opposite
Air resistance acts in the direction _____________
decreases its max. height '
to its velocity. This _____________
and range.
air
resistance
vi
gravity
no
The trajectory is _______
longer a parabola
________________________
Ex 2: A graphical example
On
way
up:
horizontal motion -________________
uniform
straight up
vertical motion –ball thrown________________
parabola
combined motion -______________
3s
2s
ay =
-9.8
m/s2
1s
vi
1s
2s
3s
coming down:
The motion is exactly the same as that of a
fired horizontally
projectile which is _______________________
:
3s
4s
5s
6s
3s
4s
5s
6s
Velocity vectors: going up
3s
2s
1s
vi
viy
vix
1s
2s
3s
vy
vx and ____
resultant velocity  found by adding ____
tangent to
 is _______________
to the parabola
vix
0 ) at the max. height.
 is = ________
(NOT = ____
Velocities coming down:
3s v
4s
5s
6s
3s
4s
5s
6s
symmetry
Notice the ______________
with going up
q on the trajectory.
The effect of changing ___
same speed
Assume all are fired with ________________
v i.
900
600
450
300
Which q results in longest range?
450
Which results in highest trajectory? 900
In longest time in air? 900
Which is a parabola? all are
y component of vi increases.
As q increases, the ___
increases
Because of this: total time in air ________________
, and
increases
maximum height ______________
angle
Complementary
________________________
angles have the same range.
compl.
angle
800
300
100
470
600
430
Range as a function
of q, assuming range
for 450 is 100. Fill
in the rest:
angle with greater….
range
neither
neither
neither
time of
flight
800
max.
height
800
600
600
470
470
100
75
50
25
0
15 30 45
60 75 90
q
Open your 3-ring binder to the
Worksheet Table of Contents.
Record the title of the worksheet:
Projectile Fired at an Angle WS
In sum: vi must be
resolved into vix and viy.
vi
viy =
visinq
q
vix = vicosq
Horizontal (x) motion:
Vertical (y) motion:
•use vix to find how far
it travels horizontally
each second
•use viy to find:
how long it’s in the air
how high it goes
•uniform x motion
•accelerated motion
• y motion is like a ball
straight up
next few slides are from an old version
Up and down together:
uniform
Horizontal motion is ____________________
at vix.
Vertical motion is like a ball tossed straight up at viy.
At top, what are v and a?
v = vix
a = -9.8 m/s2
What were v and a for the horiz. fired case at top?
v=0
a = -9.8 m/s2
How does tup compare to tdown?
equal
How does ttotal compare to tup or tdown?
ttotal=2tup=2tdown
How do the vup’s compare to vdown’s? equal
symmetry
Think: mirror _____________about
half way point.
Second half is same as _______________________
case.
horizontally fired
parabola
The shape of the trajectory is a ___________.
tangent to parabola.
Resultant v is always __________
Use the equations of motions to predict the position
and velocity of the projectile at later times.
vf = vi +at
d = vit + ½ at2
1. Since the object moves in 2 dimensions, each d, v
and a must be replaced by their components:
For x motion: d, v, a  dx, vx and ax
For y motion: d, v, a  dy, vy and ay
This is
different
vix= vicosq
from the
viy = visinq
horizontally
fired case
3. After the projectile is launched, the only
force acting on it is gravity, downward. There is no
horizontal force. Because of this, the only
acceleration a is purely vertical:
ay = -9.8 m/s2
ax = 0
2. vi is a mixture of horizontal:
and vertical:
Horizontal (x) motion:
displacement:
ax = 0
d = vit + ½ at2
dx = vixt + ½axt2
dx
dx = vixt + ½(0)t2
dx = vixt
velocity:
t
vf = vi +at
vfx = vix + axt
vfx
vfx = vix + (0)t
vfx = vix
t
The x motion is uniformat a speed = vix = vicosq
Vertical (y) motion:
displacement:
viy = 0 &
ay = -9.8 m/s2
d = vit + ½ at2
|dy|
dy = viyt + ½ayt2
dy = (0)t + ½(-9.8 m/s2)t2
dy = -4.9t2
velocity:
vf = vi +at
t
|vfy|
vfy = viy + ayt
vfy = 0 + (-9.8 m/s2)t
vfy = - 9.8t
The y motion is same as for a dropped ball.
t
Ex: A ball is fired up at an
angle as shown at right.
viy =
40
q
vix = 60 m/s
Use g = 10 m/s2
How much time does it spend going up?
4s
How much time does it spend coming down? 4 s
How much total time does it spend in the air? 8 s
What is its range?
dx = vix x ttotal = 60 m/s x 8 s
How high up does it go?
dy = viy x tup =
Find the angle q? q = tan-1(40/60)
20 m/s x 4 s
= 340
What are its v and a at maximum height?
60 m/s &
-9.8 m/s2