PROJECTILE MOTION
RULES
There are only really a few rules:
All projectiles are freefalling objects
The horizontal motion is totally unaffected
by the freefall motion!
Since there are no forces acting on it,
horizontal motion is at constant speed!
AND NOW…
HORIZONTAL PROJECTILES
•Horizontal Projectiles are easiest to work with
•only formula used in horizontal (x) direction is:
vx = d x / t
HORIZONTAL PROJECTILES
•Horizontal Projectiles are the most basic
•only formula used in horizontal (x) direction is:
vx = d x / t
constant speed!
HORIZONTAL PROJECTILES
•vertical (y) direction is just freefall
•all of the initial velocity is in the x direction
•So,
• viy = 0
vi
t1
•since viy is in
freefall,
• a = -9.8 m/s2
t2

t3


t4
EXAMPLE
A person decides to fire a rifle horizontally at a bull’s-eye.
The speed of the bullet as it leaves the barrel of the gun
is 890 m/s. He’s new to the ideas of projectile motion so
doesn’t aim high and the bullet strikes the target 1.7 cm
below the center of the bull’s-eye.
What is the horizontal distance between the rifle and the
bull’ s-eye?
start by drawing a picture:
890 m / s
1.7 cm
label the explicit givens


EXAMPLE
What is the horizontal distance between the rifle and the
bull’ s-eye?
givens (separate by direction):
X
Y
 1m 
   .017 m
d y  1.7 cm 
m 100 cm 
a y   9 .8 2
s
v iy  0 m / s
v x  890 m / s






want: dx

890 m / s

1.7 cm
horizontal distance


EXAMPLE
which equation do we use?
0
use d y  v iy t 
1
2
ay t
2
to find time

rewrite equation for t
t 
2dy
ay

2( 0.017 m )
9.8

m
s
2

0.059 s
EXAMPLE
Use t and vx to solve for dx
d x  v x t  (890

m
s
)( 0.059 s) 

52 .4 m
NON-HORIZONTAL PROJECTILES
• vx is still constant
• vy is still in freefall
•only difference with non-horizontal is…
• v iy  0
vi
NON-HORIZONTAL PROJECTILES
•Angled Projectiles require a little work to get
useful vi
•vi has an x and y component
•need to calculate initial vx and vy
vi
BREAKING UP A VECTOR
•every vector has 2 components to it
•a horizontal component
•a vertical component
vy
v
vx


•they add up to the total
BREAKING UP A VECTOR
SOHCAHTOA

sin 
hypotenuse
v
hypotenuse


sin  
O
vy

H  v
opposite

v x  v cos 
adjacent
 
 v y  v sin 
adjacent
cos 
hypotenuse
opposite
v y  v sin 
BREAKING UP A VECTOR
SOHCAHTOA
need to find θ?
hypotenuse
tan  
O
A
O 
  tan  
 A 
1
adjacent
opposite
NON-HORIZONTAL PROJECTILES
•need to calculate initial vx and vy
v y  v sin 


v
v x  v cos 
RELIEF
VISUALIZING PROJECTILES
•first enter vectors
•focus on vx
vx is constant the
whole flight!
VISUALIZING PROJECTILES
•first enter vectors
•focus on vx
•focus on vy
no vy at the top!
vy decreases as it rises!
by how much per second?
VISUALIZING PROJECTILES
VARIED ANGLES
•which projectile angle shoots highest?
•larger θ means faster viy
•which projectile angle shoots farthest?
•45° has perfect balance of fast vx
and long flight time.
LET’S ANALYZE THE JUMP
Try Graphing It
Go to pg. 331