Chapter 6B – Projectile
Motion
By
By
Professor
Professor of
of Physics
Physics
Objectives: After completing
this module, you should be
able to:
 Describe the motion of a projectile by
treating horizontal and vertical
components of its position and velocity.
• Solve for position, velocity, or time
when given initial velocity and launch
angle.
Projectile Motion
A projectile is a particle moving near
the Earth’s surface under the
influence of its weight only (directed
downward).
W
W
a=g
W
Vertical and Horizontal
Motion
Simultaneously
dropping yellow ball
and projecting red
ball horizontally.
Click right to observe
motion of each ball.
Vertical and Horizontal
Motion
Simultaneously
dropping a yellow ball
and projecting a red ball
horizontally.
W W
Why do they strike the
ground at the same
time?
Once
Once motion
motion has
has begun,
begun, the
the downward
downward
weight
weight is
is the
the only
only force
force on
on each
each ball.
ball.
Ball Projected Horizontally
and Another Dropped at
Same Time:
Vertical Motion is the Same for Each
Ball
vox
0s
vx
1s
vy
vx
vy
2s
vy
vx
vy
vy
vy
3s
Observe Motion of Each
Ball
Vertical Motion is the Same for Each
Ball
vox
0s
1s
2s
3s
Consider Horizontal and
Vertical Motion
Separately:
Compare Displacements and
Velocities
1s 2s
3s
0 s vox
vy
0s
vx
Horizontal velocity
doesn’t change.
Vertical velocity
just like free fall.
1s
vx
vy
vy
2s
vx
3s
Displacement Calculations for
Horizontal Projection:
For any constant
x vot 
acceleration:
For the special case of horizontal
projection:
a 0; a  g v 0; v v
x
y
oy
ox
o
Horizontal
displacement:
x vox t
Vertical
displacement:
y  gt
1
2
2
1
2
at
2
Velocity Calculations for
Horizontal Projection (cont.):
v f vo
For any constant
acceleration:
For the special case of a projectile:
 at
ax 0; a y  g voy 0; vox vo
Horizontal velocity:
vx vox
Vertical velocity:
v y vo  gt
Example 1: A baseball is hit with a
horizontal speed of 25 m/s. What
is its position and velocity after 2
s?
x +50 m
25 m/s
y
-19.6 m
First find horizontal and vertical
displacements:
x vox t (25 m/s)(2 s)
xx =
= 50.0
50.0
m
m
2
2
2
1
1
y  2 gt  2 ( 9.8 m/s )(2 s) yy =
= -19.6
-19.6
m
m
Example 1 Cont.): What are the
velocity components after 2 s?
25 m/s
v0x = 25 m/s
v0y = 0
vx
vy
Find horizontal and vertical velocity after 2 s:
vx vox (25 m/s)
vvxx =
= 25.0
25.0
m/s
2m/s
v y voy  at 0  ( 9.8 m/s )(2 s)
vvyy =
= -19.6
-19.6
Consider Projectile at an
Angle:
AA red
red ball
ball is
is projected
projected at
at an
an angle
angle .. At
At the
the
same
same time,
time, aa yellow
yellow ball
ball is
is thrown
thrown
vertically
vertically upward
upward and
and aa green
green ball
ball rolls
rolls
horizontally
horizontally (no
(no friction).
friction).
vx = vox = constant
vo
voy

vox
v y voy  at
a  9.8 m/s
2
Note vertical and horizontal motions of
balls
Displacement Calculations For
General Projection:
The components of displacement at time t
are:
2
22
xx 
vvoxoxtt  aaxxtt
11
22
2
yy 
v
t

a
t
voyoy t  ayy t
11
22
For
ax 0; a y  g ; voy 0; vox vo
projectiles:
Thus, the
x vox t
displacement
2
1
y voy t  2 gt
components x and y
for projectiles are:
Velocity Calculations For
General Projection:
The components of velocity at time t
are:
vvxx 
vvoxox aaxxtt
For
projectiles:
vvyy 
vvoyoy aayytt
ax 0; a y  g ; voy 0; vox vo
Thus, the velocity
components vx and vy
for projectiles are:
vx vox constant
v y voy  gt
Problem-Solving Strategy:
1. Resolve initial velocity vo into
components:
vo
v
oy

vox
vvoxox 
vvoo cos
cos ;;
vvoyoy 
vvoo sin
sin
2. Find components of final position and
velocity:
Displacement:
Velocity:
vx v0 x
x vox t
y voy t  gt
1
2
2
v y voy  gt
2
Problem Strategy (Cont.):
3. The final position and velocity can be
found from the components.
y
voy
R

x
yy
RR 
 xx  yy ;; tan
tan 

xx
vo

vox
vvyy
vv 
 vv vv ;; tan
tan 

vvxx
22
22
xx
22
22
yy
4. Use correct signs - remember g is
negative or positive depending on your
initial choice.
Example 2: A ball has an initial
velocity of 160 ft/s at an angle of 30o
with horizontal. Find its position and
velocity after 2 s and after 4 s.
voy 160 ft/s v (160 ft/s) cos 300 139 ft/s
ox
30ov
0
ox
voy (160 ft/s) sin 30 80.0 ft/s
Since vx is constant, the horizontal
displacements after 2 and 4 seconds are:
x vox t (139 ft/s)(2 s)
xx =
= 277
277 ft
ft
x vox t (139 ft/s)(4 s)
xx =
= 554
554 ft
ft
Example 2: (Continued)
voy 160 ft/s
30ov
2
s
ox
277 ft
4
s
554
ft
Note: We know ONLY the horizontal
location after 2 and 4 s. We don’t know
whether it is on its way up or on its way
down.
xx22 =
= 277
277 ft
ft
xx44 =
= 554
554 ft
ft
Example 2 (Cont.): Next we find the
vertical components of position after 2 s
and after 4 s.
g = -32 ft/s2
voy= 80 ft/s
160 ft/s
y2

y4
0
1
2
3
4
s
s
s
s
s
The vertical displacement as function of
time:
2
2 2
1
1
y voy t  2 gt (80 ft/s)t  2 ( 32 ft/s )t
22
yy 
80
80tt  16
16tt
Observe consistent
units.
(Cont.) Signs of y will indicate
location of displacement (above + or
below – origin).
g = -32 ft/s2
voy= 80 ft/s
160 ft/s
96 ft
y2
16 ft

0
s
1
2
s
s
Vertical
position:
y2 80(2 s)  16(2 s)
yy22 
96
96 ftft
3
s
y4
4
s
22
yy 
80
80tt  16
16tt
2
yy44 
16
16 ftft
y4 80(4 s)  16(4 s)
2
Each above origin
(+)
(Cont.): Next we find horizontal and
vertical components of velocity after 2
and 4 s.
voy 160 ft/s v (160 ft/s) cos 300 139 ft/s
ox
30ov
0
ox
voy (160 ft/s) sin 30 80.0 ft/s
Since
Since vvxx is
is constant,
constant, vvxx =
= 139
139 ft/s
ft/s at
at all
all
times.
times.
Vertical velocity is same as if vertically
projected:
2
v y voy  gt ;
At any
time t:
where
vx 139 ft/s
g  32 ft/s
v y voy  (32 ft/s)t
Example 2: (Continued)
g = -32 ft/s2
v2
vy= 80.0
ft/s
160 ft/s
v4

0
s
At any
time t:
1
s
2
s
vx 139 ft/s
3
s
4
s
v y voy  (32 ft/s)t
v y 80 ft/s  (32 ft/s)(2 s)
vv2y
= 16.0 ft/s
2y = 16.0 ft/s
v y 80 ft/s  (32 ft/s)(4 s) vv4y =
-48.0
ft/s
=
-48.0
ft/s
4y
Example 2: (Continued)
g = -32 ft/s2
v2
vy= 80.0
ft/s
160 ft/s

Moving
Up +16
ft/s
Moving
down -48
ft/s
v4
0
1
2
3
4
s
s
s
s
s
The signs of vy indicate whether motion
is up (+) or down (-) at any time t.
At
At 22 s:
s:
ft/s
ft/s
At
At 44 s:
s:
vv2x
= 139 ft/s; v 2y =
=+
+ 16.0
16.0
2x = 139 ft/s; v2y
vv4x
=
139
ft/s;
v
=
48.0
ft/s
=
139
ft/s;
v
=
48.0
ft/s
4y
4x
4y
(Cont.): The displacement R2, is found
from the x2 and y2 component
displacements.
2
R x y
2
R2
t=2
s
y2 = 96 ft
y
tan  
x

0
s
x2= 277
ft
2
s
R  (277 ft) 2  (96 ft) 2
RR22 =
= 293
293
ft
ft
4
s
96 ft
tan  
277 ft
00
 =
19.1
= 19.1
(Cont.): Similarly, displacement R4, is
found from the x4 and y4 component
displacements.
2
R x y
y
tan  
x
t=4
s
R4
2
y4 = 64

0
s
x4= 554
ft
2
R  (554 ft)  (64 ft)
RR44 =
= 558
558
ft
ft
4
s
2
ft
64 ft
tan  
554 ft
00
 =
6.59
= 6.59
(Cont.): Now we find the velocity
after 2 s from the components vx
and vy.
v2
g = -32 ft/s2
voy= 80.0
ft/s
160 ft/s
v2x = 139 ft/s
v2y = + 16.0 ft/s
Moving
Up +16
ft/s

0
s
2
s
2
v2  (139 ft/s)  (16 ft/s)
vv22 =
= 140
140 ft/s
ft/s
2
16 ft
tan  
139 ft
 =
= 6.56
6.5600
(Cont.) Next, we find the velocity
after 4 s from the components v4x
and v4y.
g = -32 ft/s2
voy= 80.0
ft/s
160 ft/s

v4x = 139 ft/s
v4y = - 48.0 ft/s
0
s
v4  (139 ft/s) 2  (  46 ft/s) 2
vv44 =
= 146
146 ft/s
ft/s
v4
4
s
16 ft
tan  
139 ft
00
 =
341.7
= 341.7
Example 3: What are maximum height
and range of a projectile if vo = 28 m/s
at 300?
voy 28 m/s
30ov
vy = 0
vox = 24.2 m/s
ymax v = + 14 m/s
oy
ox
0
vox (28 m/s) cos 30 24.2 m/s
0
voy (28 m/s) sin 30 14 m/s
Maximum y-coordinate occurs when vy =
0: v y voy  gt 14 m/s  ( 9.8 m/s 2 )t 0
ymax occurs when 14 – 9.8t = 0 or t =
Example 3(Cont.): What is maximum
height of the projectile if v = 28 m/s at
300?
voy 28 m/s
30ov
vy = 0
vox = 24.2 m/s
ymax v = + 14 m/s
oy
ox
Maximum y-coordinate occurs when t =
1.43 s: 1 2
2
1
y voy t  2 gt 14(1.43)  2 ( 9.8)(1.43)
y 20 m  10 m
yymax
= 10.0
max= 10.0
m
m
Example 3(Cont.): Next, we find the
range of the projectile if v = 28 m/s at
300.
voy 28 m/s
30o
vox = 24.2 m/s
voy = + 14 m/s
vox
Range xr
The range xr is defined as horizontal
distance coinciding with the time for
vertical
The timereturn.
of flight is found by setting y
=2 0:
y voy t  gt 0
1
2
(continued
)
Example 3(Cont.): First we find the
time of flight tr, then the range xr.
voy 28 m/s
30o
vox = 24.2 m/s
voy = + 14 m/s
vox
Range xr
2
y voy t  gt 0
1
2
voy  gt 0;
1
2
(Divide by t)
voy
2(14 m/s)
t

; t 2.86 s
2
 g -(-9.8 m/s )
xr = voxt = (24.2 m/s)(2.86
s);
xxrr =
= 69.2
69.2
m
m
Example 4: A ball rolls off the top
of a table 1.2 m high and lands on
the floor at a horizontal distance
of 2 m. What was the velocity as it
left the table?
R
1.2 m
2m
First find t from y
equation:
½(-9.8)t2 = (1.2)
Note: x = voxt = 2 m
0
y = voyt + ½ayt2 = -1.2 m
2
y  gt  1.2 m
1
2
2( 1.2)
t
 9.8
tt =
= 0.495
0.495
Example 4 (Cont.): We now use
horizontal equation to find vox leaving
the table top.
R
1.2 m
Note: x = voxt = 2 m
2m
Use t = 0.495 s in x
equation:
v ox (0.495 s) = 2 m;
The ball leaves
the table with a
speed:
y = ½gt2 = -1.2 m
vox t 2 m
2m
vox 
0.495 s
vv =
= 4.04
4.04
m/s
m/s
Example 4 (Cont.): What will be its
speed when it strikes the floor?
1.2 m
Note:
t = 0.495
s
vx
2m
vy
vy = 0 + (-9.8 m/s2)(0.495
s)
2
v  (4.04 m/s)  (  4.85 m/s)
vv44 =
= 146
146 ft/s
ft/s
vx = vox = 4.04
0
m/s
v = v + gt
y
2
y
vy = -4.85
m/s
 4.85 m
tan  
4.04 m
00
 =
309.8
= 309.8
Example 5. Find the “hang time” for the
football whose initial velocity is 25 m/s,
600.
y = 0; a = -9.8
vo =25 m/s
m/s2
Time of
0
60
flight t
Initial vo:
vox = vo cos 
voy = vo sin 
Vox = (25 m/s) cos 600; vox = 12.5
m/s 0
Voy = (25 m/s) sin 60 ; vox = 21.7
m/s
Only vertical parameters affect hang
time.
2
2
1
1
y voy t  2 at ; 0 (21.7)t  2 ( 9.8)t
Example 5 (Cont.) Find the “hang time” for
the football whose initial velocity is 25 m/s,
600.
y = 0; a = -9.8
vo =25 m/s
m/s2
Time of
0
60
flight t
2
Initial vo:
vox = vo cos 
voy = vo sin 
y voy t  at ; 0 (21.7)t  ( 9.8)t
1
2
4.9 t2 = 21.7 t
21.7 m/s
t
4.9 m/s 2
1
2
4.9 t = 21.7
tt =
= 4.42
4.42 ss
2
Example 6. A running dog leaps with
initial velocity of 11 m/s at 300. What is
the range?
Draw figure and
voy = 11 sin
find
300
v = 11
components:
m/s
vox = 9.53 m/s
0
=30
voy = 5.50 m/s
vox = 11 cos
300
To find range, first find t when y = 0; a = -9.8
2
2
m/s2
1
1
y v t  at ; 0 (5.50)t  ( 9.8)t
oy
2
4.9 t2 = 5.50 t
4.9 t = 5.50
2
5.50 m/s
t
2
4.9 m/s
tt =
= 1.12
1.12 ss
Example 6 (Cont.) A dog leaps with
initial velocity of 11 m/s at 300. What is
the range?
Range is found
voy = 10 sin
from x310
v = 10
component:
m/s
vx = vox = 9.53

=310
m/s
vox = 10 cos
x = vxt; t = 1.12
0
31
s
Horizontal velocity is constant: vx = 9.53 m/s
x = (9.53 m/s)(1.12 s) = 10.7 m
Range:
Range: xx =
= 10.7
10.7
m
m
Summary for Projectiles:
1. Determine x and y components v0
vvoxox 
vvoo cos
cos and
and
vvoyoy 
vvoosin
sin
2. The horizontal and vertical
components of displacement at any
time t are given by:
xx 
vvoxoxtt
22
yy 
vvoyoytt  gt
gt
11
22
Summary (Continued):
3. The horizontal and vertical
components of velocity at any time t
are given by:
vvxx 
vvoxox;; vvyy 
vvoyoy  gt
gt
4. Vector displacement or velocity can
then be found from the components
if desired:
yy
22
22
tan
R
tan 

R
 xx  yy
xx
CONCLUSION: Chapter 6B
Projectile Motion