Non-Linear Motion: Projectile Motion

Contributed by:

Jonathan James Thu, Jan 20, 2022 06:53 PM UTC

2-D Projectile, Horizontally launched projectiles, Vertically launched projectiles

1. Projectile Motion
AP Physics B

2. What is projectile?
Projectile -Any object which projected by some
means and continues to move due to its own
inertia (mass).

3. Projectiles move in TWO
Since a projectile
moves in 2-
dimensions, it
therefore has 2
components just
like a resultant
vector.
 Horizontal and
Vertical

4. Horizontal “Velocity”
 NEVER changes, covers equal displacements in
equal time periods. This means the initial
horizontal velocity equals the final horizontal
velocity
In other words, the horizontal
velocity is CONSTANT. BUT
WHY?
Gravity DOES NOT work
horizontally to increase or
decrease the velocity.

5. Vertical “Velocity”
 Changes (due to gravity), does NOT cover
equal displacements in equal time periods.
Both the MAGNITUDE and DIRECTION change. As
the projectile moves up the MAGNITUDE
DECREASES and its direction is UPWARD. As it
moves down the MAGNITUDE INCREASES and the
direction is DOWNWARD.

6. Combining the Components
Together, these
components produce
what is called a
trajectory or path. This
path is parabolic in
nature.
Component Magnitude Direction
Horizontal Constant Constant
Vertical Changes Changes

7. Horizontally Launched
Projectiles
Projectiles which have NO upward trajectory and NO initial
VERTICAL velocity. vox vx constant
voy 0 m / s

8. Horizontally Launched
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for
the “y” direction. And for this we use kinematic #2.
1
x voxt  at 2
2
x vox t y  1 gt 2
2
Remember, the velocity is Remember that since the
CONSTANT horizontally, so projectile is launched
that means the acceleration horizontally, the INITIAL
is ZERO! VERTICAL VELOCITY is
equal to ZERO.

9. Horizontally Launched
Example: A plane traveling with What do I What I want to
a horizontal velocity of 100 know? know?
m/s is 500 m above the
ground. At some point the vox=100 m/s t=?
pilot drops a bomb on a
target below. (a) How long is y = 500 m x=?
the bomb in the air? (b) How
far away from point above voy= 0 m/s
where it was dropped will it
land? g = -9.8 m/s/s
y  1 gt 2   500  1 ( 9.8)t 2
2 2
x vox t (100)(10.1)  1010 m
102.04 t 2  t  10.1 seconds

10. Vertically Launched
NO Vertical Velocity at the top of the trajectory.
Vertical Vertical Velocity
Velocity increases on the
decreases way down,
on the way
upward Horizontal Velocity
is constant
Component Magnitude Direction
Horizontal Constant Constant
Vertical Decreases up, 0 Changes
@ top, Increases
down

11. Vertically Launched
Since the projectile was launched at a angle, the
velocity MUST be broken into components!!!
vox vo cos 
vo voy
voy vo sin 

vox

12. Vertically Launched
There are several
things you must
consider when doing
these types of
projectiles besides
using components. If
it begins and ends at
ground level, the “y”
displacement is
ZERO: y = 0

13. Vertically Launched
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
voy x voxt y voy t  1 gt 2
vo 2

vox vox vo cos 
voy vo sin 

14. A place kicker kicks a football with a velocity of 20.0 m/s
and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
vox vo cos 
/s
m
0.0 vox 20 cos 53 12.04 m / s
=2
vo
voy vo sin 

voy 20sin 53 15.97 m / s

15. A place kicker kicks a What I know What I want
football with a to know
velocity of 20.0 m/s vox=12.04 m/s t=?
and at an angle of 53 voy=15.97 m/s x=?
degrees.
y=0 ymax=?
(a) How long is the ball
in the air? g = - 9.8
m/s/s
y voy t  1 gt 2  0 (15.97)t  4.9t 2
2
2
 15.97t  4.9t  15.97 4.9t
t  3.26 s

16. A place kicker kicks a What I know What I want
football with a to know
velocity of 20.0 m/s vox=12.04 m/s t = 3.26 s
and at an angle of 53 voy=15.97 m/s x=?
degrees. y=0 ymax=?
(b) How far away does it
g = - 9.8
land? m/s/s
x vox t  (12.04)(3.26)  39.24 m

17. Example What I know What I want
to know
A place kicker kicks a vox=12.04 m/s t = 3.26 s
football with a velocity voy=15.97 m/s x = 39.24 m
of 20.0 m/s and at an
angle of 53 degrees. y=0 ymax=?
g = - 9.8
(c) How high does it m/s/s
travel?
y voy t  1 gt 2
2
CUT YOUR TIME IN HALF! y (15.97)(1.63)  4.9(1.63) 2
y  13.01 m

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