Chapter 6B – Projectile Motion
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
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 motion has begun, the downward
weight is the only force on each ball.
Ball Projected Horizontally and
Another Dropped at Same Time:
Vertical Motion is the Same for Each Ball
vox
0s
vx
vy
vx
vy
vy
vx
vy
vy
vy
1s
2s
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
0s
vox 1 s
vy
2s
3s
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 acceleration:
x  vot  at
1
2
For the special case of horizontal projection:
ax  0; a y  g voy  0; vox  vo
Horizontal displacement:
x  voxt
Vertical displacement:
y  12 gt 2
2
Velocity Calculations for Horizontal
Projection (cont.):
For any constant acceleration:
v f  vo  at
For the special case of a projectile:
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
y
25 m/s
-19.6 m
First find horizontal and vertical displacements:
x  voxt  (25 m/s)(2 s)
y  gt  (9.8 m/s )(2 s)
1
2
2
1
2
2
x = 50.0 m
2
y = -19.6 m
Example 1 Cont.): What are the velocity
components after 2 s?
25 m/s
vx
v0x = 25 m/s
v0y = 0
vy
Find horizontal and vertical velocity after 2 s:
vx  vox  (25 m/s)
vx = 25.0 m/s
vy  voy  at  0  (9.8 m/s )(2 s)
2
vy = -19.6 m/s
Consider Projectile at an Angle:
A red ball is projected at an angle q. At the same
time, a yellow ball is thrown vertically upward
and a green ball rolls horizontally (no friction).
vx = vox = constant
vo
voy
q
v y  voy  at
vox
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:
x  vox t  axt
1
2
2
y  voyt  ayt
1
2
2
For projectiles: ax  0; a y  g ; voy  0; vox  vo
Thus, the displacement
components x and y for
projectiles are:
x  voxt
y  voy t  gt
1
2
2
Velocity Calculations For
General Projection:
The components of velocity at time t are:
vx  vox  axt
v y  voy  a y t
For projectiles: 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:
voy
vo
q
vox
vox  vo cos q ;
voy  vo sin q
2. Find components of final position and velocity:
Displacement:
Velocity:
vx  v0 x
x  voxt
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
q
x
R
vo
q
vox
v  v v ;
x y ;
2
2
x
2
2
y
y
tan q 
x
tan q 
vy
vx
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
vox  (160 ft/s) cos 30  139 ft/s
30o
voy  (160 ft/s)sin 30  80.0 ft/s
vox
0
0
Since vx is constant, the horizontal displacements
after 2 and 4 seconds are:
x  voxt  (139 ft/s)(2 s)
x = 277 ft
x  voxt  (139 ft/s)(4 s)
x = 554 ft
Example 2: (Continued)
voy
160 ft/s
30o
vox
2s
277 ft
4s
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.
x2 = 277 ft
x4 = 554 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
q
0s
1s
2s
3s
4s
The vertical displacement as function of time:
y  voyt  gt  (80 ft/s)t  (32 ft/s )t
1
2
2
y  80t  16t
1
2
2
2
2
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
q
0s
1s
2s
3s
Vertical position:
y2  80(2 s)  16(2 s)
y2  96 ft
2
y4  16 ft
y4
4s
y  80t  16t
2
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
vox  (160 ft/s) cos 30  139 ft/s
30o
voy  (160 ft/s)sin 30  80.0 ft/s
vox
0
0
Since vx is constant, vx = 139 ft/s at all times.
Vertical velocity is same as if vertically projected:
vy  voy  gt;
At any
time t:
where
vx  139 ft/s
g  32 ft/s
2
v y  voy  (32 ft/s)t
Example 2: (Continued)
vy= 80.0 ft/s
g = -32 ft/s2
v2
160 ft/s
v4
q
0s
At any
time t:
1s
2s
vx  139 ft/s
3s
4s
v y  voy  (32 ft/s)t
v y  80 ft/s  (32 ft/s)(2 s)
v2y = 16.0 ft/s
v y  80 ft/s  (32 ft/s)(4 s)
v4y = -48.0 ft/s
Example 2: (Continued)
2
g
=
-32
ft/s
vy= 80.0 ft/s
v2
160 ft/s
Moving Up
+16 ft/s
Moving down
-48 ft/s
q
0s
1s
2s
3s
v4
4s
The signs of vy indicate whether motion is
up (+) or down (-) at any time t.
At 2 s: v2x = 139 ft/s; v2y = + 16.0 ft/s
At 4 s: v4x = 139 ft/s; v4y = - 48.0 ft/s
(Cont.): The displacement R2,q is found from
the x2 and y2 component displacements.
R
x y
2
2
R2
t=2s
y2 = 96 ft
y
tan q 
x
q
0 s x2= 277 ft 2 s
R  (277 ft)  (96 ft)
2
R2 = 293 ft
4s
2
96 ft
tan q 
277 ft
q2 = 19.10
(Cont.): Similarly, displacement R4,q is found
from the x4 and y4 component displacements.
R
x y
2
t=4s
2
R4
y4 = 64 ft
q
0s
x4= 554 ft
R  (554 ft)  (64 ft)
2
R4 = 558 ft
y
tan q 
x
2
4s
64 ft
tan q 
554 ft
q4 = 6.590
(Cont.): Now we find the velocity after
2 s from the components vx and vy.
2
v
g
=
-32
ft/s
2
v = 80.0 ft/s
oy
v2x = 139 ft/s
v2y = + 16.0 ft/s
Moving Up
+16 ft/s
160 ft/s
q
0s
2s
v2  (139 ft/s)  (16 ft/s)
2
v2 = 140 ft/s
2
16 ft
tan q 
139 ft
q2 = 6.560
(Cont.) Next, we find the velocity after
4 s from the components v4x and v4y.
g = -32 ft/s2
voy= 80.0 ft/s
v4x = 139 ft/s
v4y = - 48.0 ft/s
160 ft/s
q
0s
v4
4s
v4  (139 ft/s)  (46 ft/s)
2
v4 = 146 ft/s
2
16 ft
tan q 
139 ft
q2 = 341.70
Example 3: What are maximum height and
range of a projectile if vo = 28 m/s at 300?
voy
28 m/s
30o
vy = 0
ymax
vox
vox = 24.2 m/s
voy = + 14 m/s
vox  (28 m/s) cos 30  24.2 m/s
0
voy  (28 m/s)sin 30  14 m/s
0
Maximum y-coordinate occurs when vy = 0:
vy  voy  gt  14 m/s  (9.8 m/s )t  0
2
ymax occurs when 14 – 9.8t = 0 or t = 1.43 s
Example 3(Cont.): What is maximum height
of the projectile if v = 28 m/s at 300?
voy
28 m/s
30o
vy = 0
vox
ymax
vox = 24.2 m/s
voy = + 14 m/s
Maximum y-coordinate occurs when t = 1.43 s:
y  voyt  gt  14(1.43)  (9.8)(1.43)
1
2
2
y  20 m  10 m
1
2
ymax= 10.0 m
2
Example 3(Cont.): Next, we find the range
of the projectile if v = 28 m/s at 300.
voy
vox = 24.2 m/s
voy = + 14 m/s
28 m/s
30o
vox
Range xr
The range xr is defined as horizontal distance
coinciding with the time for vertical return.
The time of flight is found by setting y = 0:
y  voyt  gt  0
1
2
2
(continued)
Example 3(Cont.): First we find the time of
flight tr, then the range xr.
voy
vox = 24.2 m/s
voy = + 14 m/s
28 m/s
30o
vox
Range xr
y  voyt  gt  0
1
2
voy  gt  0;
1
2
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);
xr = 69.2 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?
1.2 m
R
2m
Note: x = voxt = 2 m
0
y = voyt + ½ayt2 = -1.2 m
First find t from y equation:
½(-9.8)t2 = -(1.2)
y  gt  1.2 m
1
2
2
2(1.2)
t
9.8
t = 0.495 s
Example 4 (Cont.): We now use horizontal
equation to find vox leaving the table top.
1.2 m
R
Note: x = voxt = 2 m
2m
y = ½gt2 = -1.2 m
Use t = 0.495 s in x equation:
v ox (0.495 s) = 2 m;
The ball leaves the
table with a speed:
voxt  2 m
2m
vox 
0.495 s
v = 4.04 m/s
Example 4 (Cont.): What will be its speed
when it strikes the floor?
Note:
t = 0.495 s
vx = vox = 4.04 m/s
1.2 m
0
vx
2m
vy = vy + gt
vy
vy = -4.85 m/s
vy = 0 + (-9.8 m/s2)(0.495 s)
v  (4.04 m/s)  (4.85 m/s)
2
v4 = 146 ft/s
2
4.85 m
tan q 
4.04 m
q2 = 309.80
Example 5. Find the “hang time” for the football
whose initial velocity is 25 m/s, 600.
vo =25 m/s
600
Initial vo:
y = 0; a = -9.8 m/s2
Time of
flight t
vox = vo cos q
voy = vo sin q
Vox = (25 m/s) cos 600;
vox = 12.5 m/s
Voy = (25 m/s) sin 600;
vox = 21.7 m/s
Only vertical parameters affect hang time.
y  voyt  at ; 0  (21.7)t  (9.8)t
1
2
2
1
2
2
Example 5 (Cont.) Find the “hang time” for the
football whose initial velocity is 25 m/s, 600.
vo =25 m/s
600
Initial vo:
y = 0; a = -9.8 m/s2
Time of
flight t
vox = vo cos q
voy = vo sin q
y  voyt  at ; 0  (21.7)t  (9.8)t
1
2
2
4.9 t2 = 21.7 t
21.7 m/s
t
4.9 m/s 2
1
2
4.9 t = 21.7
t = 4.42 s
2
Example 6. A running dog leaps with initial
velocity of 11 m/s at 300. What is the range?
Draw figure and
find components:
voy = 11 sin 300
v = 11 m/s
vox = 9.53 m/s
voy = 5.50 m/s
q =300
vox = 11 cos 300
To find range, first find t when y = 0; a = -9.8 m/s2
y  voyt  at ; 0  (5.50)t  (9.8)t
1
2
4.9 t2 = 5.50 t
4.9 t = 5.50
2
1
2
5.50 m/s
t
2
4.9 m/s
2
t = 1.12 s
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 310
from x-component:
v = 10 m/s
vx = vox = 9.53 m/s
x = vxt; t = 1.12 s
q =310
vox = 10 cos 310
Horizontal velocity is constant: vx = 9.53 m/s
x = (9.53 m/s)(1.12 s) = 10.7 m
Range: x = 10.7 m
Summary for Projectiles:
1. Determine x and y components v0
vox  vo cos q
and
voy  vo sin q
2. The horizontal and vertical components of
displacement at any time t are given by:
x  voxt
y  voyt  gt
1
2
2
Summary (Continued):
3. The horizontal and vertical components of
velocity at any time t are given by:
vx  vox ; v y  voy  gt
4. Vector displacement or velocity can then
be found from the components if desired:
R
x y
2
2
y
tan q 
x
CONCLUSION: Chapter 6B
Projectile Motion