3. up at an angle q
Projectile fired __________________________
with an initial _________________
Assume no _________________. The only force
acting on the projectile is _________ . This means
the acceleration is ____________, ______________
speed vi
air resistance
gravity
-9.81 m/s2
downward
vtop ______, atop = ___________
≠ 0
-9.81 m/s2
The velocity
is always
__________
the path
tangent to
gravity vi v
max. height q
the range

4. To solve the problem,
vi must be ____________
into its horizontal (vix)
and vertical (viy)
_____________________. vi
resolved
viy =
_______
visinq
components q
vix =____________
vicosq
Where: vi = _______________ is the initial speed,
and q = __________________ is the angle.
√(vix2 + viy2)
tan-1(viy/vix)
There are _____ simultaneous motions:
For ___ motion, use: _____________________ 2 x
dx, vx and ax
dy, vy and ay y

5. vix
The horizontal motion is determined by ___ =
_______ . Because there is _______ horizontal force,
vix __________________  _____________ x-motion.
vicosq no
remains constant
uniform
acceleration:
ax = ax t
velocity:
vfx =
vix
vfx t
vixt dx
displacement:
dx = t

6. B. Vertical motion is determined by ___ = _______ .
viy
B. Vertical motion is determined by ___ = _______ .
Because of ____________, the y motion is like a ball
thrown _______________ with an initial speed ____ .
visinq
gravity
straight up
viy
acceleration:
ay = -g ay =
-9.81 m/s2 -g t
velocity:
vfy =
viy + ayt
viy
vfy
= visinq – 9.81 m/s2 t t
displacement:
dy =
viyt + (1/2)ayt2 dy
= visinqt – (1/2)9.81 m/s2 t2 t

7. Ex 1: Mr. Siudy is fired out of a cannon at a speed
of 75 m/s and at an angle of 370 to the horizontal.
vix = vicosq
= (75 m/s)cos370
75 m/s
viy =
60. m/s
370
viy = visinq
= (75 m/s)sin370
vix =
45. m/s
To determine how high up he goes and how long
he is in the air, "pretend" he is fired __________
___ but with an initial speed = _____ = __________
straight
viy up
45. m/s
viy =
Given:
45. m/s
1st Unknown: dy
ay =
-9.8 m/s2
2nd Unknown: t
vfy =

8. How far up?
vf2 = vi2 + 2ad
02 = (-9.81 m/s2)d
103 m = d
How long is he in the air?
vf = vi + at
0 = 45+ (-9.81 m/s2)t
4.6 s = t
Because we chose vfy = ___ , this t represents the
time to _________________ . To get the total time
of flight, we must _____________________ . So, the
total time t = _______ s. You could get this time
directly if you assume vfy = __________ . Then:
rise only
double this time
9.2
-45 m/s
vf = vi + at

9. -45 = m/s2 (t)
9.2 s = t
To determine his range, you must assume his
x motion is ____________ at vi = ____ = _______ .
uniform
vix
60. m/s
Given:
vix =
60. m/s
Unknown: dx
ax =
t =
9.2 s
total
Notice that the ___________ time is used here!
d = vit + (1/2)at2 =
(60.)9.2 s + =
550 m

10. With no ____________________, only the force of
___________ acts on the object:
air resistance
gravity
The
trajectory (path)
is a________________. vi
gravity
parabola
Air resistance acts in the direction _____________
to its velocity. This _____________ its max. height '
and range.
opposite
decreases
The trajectory is _______
________________________ no
air
resistance
longer a parabola vi
gravity

11. Ex 2: A graphical exampleOn
way
up:
horizontal motion -________________
uniform
vertical motion –ball thrown________________
straight up
combined motion -______________
parabola
3 s
2 s
ay =
1 s
-9.81
m/s2 vi
1 s
2 s
3 s

12. The motion is exactly the same as that of a
coming down:
The motion is exactly the same as that of a
projectile which is _______________________ :
fired horizontally
3 s
4 s
5 s
6 s
4 s
5 s
6 s
3 s

13. Velocity vectors: going upvi
viy
vix
1 s
2 s
3 s vx vy
resultant velocity  found by adding ____ and ____
 is _______________ to the parabola
 is = ________ (NOT = ____ ) at the max. height.
tangent to
vix

14. Velocities coming down:
symmetry
Notice the ______________ with going up

15. The effect of changing ___ on the trajectory.
Assume all are fired with ________________ vi. q
same speed
900
600
450
300
Which q results in longest range?
Which results in highest trajectory?
In longest time in air?
Which is a parabola?
450
900
900
all are

16. As q increases, the ___ component of vi increases.
Because of this: total time in air ________________ , and
maximum height ______________ y
increases
increases
Complementary
________________________ angles have the same range.
compl.
angle
angle with greater….
range
time of
flight
max.
height
800
600
470
angle
neither
800
800
100
300
neither
600
600
430
neither
470
470 25 15 30 45 60 75 90 q 50
100
Range as a function
of q, assuming range
for 450 is 100. Fill
in the rest:

17. vix = vicosq
viy =
visinq q
In sum: vi must be
resolved into vix and viy. vi
Vertical (y) motion:
use viy to find:
how long it’s in the air
how high it goes
accelerated motion
y motion is like a ball
straight up
Horizontal (x) motion:
use vix to find how far
it travels horizontally
each second
uniform x motion

18. uniform Up and down together:
Horizontal motion is ____________________ at vix.
Vertical motion is like a ball tossed straight up at viy.

19. At top, what are v and a?
What were v and a for the horiz. fired case at top?
How does tup compare to tdown?
How does ttotal compare to tup or tdown?
How do the vup’s compare to vdown’s?
Think: mirror _____________about half way point.
Second half is same as _______________________ case.
The shape of the trajectory is a ___________.
Resultant v is always __________ to parabola.
v = vix a = m/s2
v = a = m/s2
equal
ttotal=2tup=2tdown
equal
symmetry
horizontally fired
parabola
tangent

20. Use the equations of motions to predict the position
and velocity of the projectile at later times.
d = vit + ½ at2
vf = vi +at
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
from the
horizontally
fired case
vi is a mixture of horizontal: vix= vicosq
and vertical: viy = visinq
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 = m/s2
ax = 0

21. Horizontal (x) motion: ax = 0
displacement:
d = vit + ½ at2
dx = vixt + ½axt2 dx
dx = vixt + ½(0)t2
dx = vixt t
velocity:
vf = vi +at
vfx = vix + axt
vfx
vfx = vix + (0)t t
vfx = vix
The x motion is uniformat a speed =
vix = vicosq

22. Vertical (y) motion: viy = 0 & ay = -9.81 m/s2
displacement:
d = vit + ½ at2
|dy|
dy = viyt + ½ayt2
dy = (0)t + ½(-9.81 m/s2)t2
dy = -4.9t2 t
velocity:
vf = vi +at
|vfy|
vfy = viy + ayt
vfy = 0 + (-9.81 m/s2)t t
vfy = - 9.8t
The y motion is same as for a dropped ball.

23. Ex: A ball is fired up at an
angle as shown at right.
viy = 40 q
Use g = 10 m/s2
vix = 60 m/s
How much time does it spend going up?
4 s
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 =
20 m/s x 4 s
Find the angle q?
q = tan-1(40/60)
= 340
What are its v and a at maximum height?
60 m/s &
-9.8 m/s2