2. 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.
3. Projectile Motion
A projectile is a particle moving near the
Earth’s surface under the influence of its
weight only (directed downward).
a = g
W
W
W
4. Vertical and Horizontal Motion
Simultaneously dropping
yellow ball and projecting
red ball horizontally.
Click right to observe
motion of each ball.
5. Vertical and Horizontal Motion
Simultaneously dropping a
yellow ball and projecting a
red ball horizontally.
Why do they strike the
ground at the same time?
Once motion has begun, the downward
weight is the only force on each ball.
W W
6. Ball Projected Horizontally and
Another Dropped at Same Time:
0 s
vox
Vertical Motion is the Same for Each Ball
1 s
2 s
3 s
vy
vx
vx
vx
vy
vy
vy
vy
vy
7. Observe Motion of Each Ball
0 s
vox
Vertical Motion is the Same for Each Ball
3 s
2 s
1 s
8. Consider Horizontal and
Vertical Motion Separately:
Compare Displacements and Velocities
0 s
0 s
1 s
vox
2 s 3 s
1 s
vy
2 s
vx
vy
3 s
vx
vy
Horizontal velocity
doesn’t change.
Vertical velocity just
like free fall.
vx
9. Displacement Calculations for
Horizontal Projection:
For any constant acceleration:
Horizontal displacement: ox
x v t
Vertical displacement:
2
1
2
y gt
2
1
2
o
x v t at
For the special case of horizontal projection:
0; 0;
x y oy ox o
a a g v v v
10. Velocity Calculations for Horizontal
Projection (cont.):
For any constant acceleration:
Horizontal velocity: x ox
v v
Vertical velocity: y o
v v gt
f o
v v at
For the special case of a projectile:
0; 0;
x y oy ox o
a a g v v v
11. Example 1: A baseball is hit with a
horizontal speed of 25 m/s. What is its
position and velocity after 2 s?
First find horizontal and vertical displacements:
(25 m/s)(2 s)
ox
x v t
2 2 2
1 1
2 2 ( 9.8 m/s )(2 s)
y gt
x = 50.0 m
y = -19.6 m
25 m/s
x
y
-19.6 m
+50 m
12. Example 1 Cont.): What are the velocity
components after 2 s?
25 m/s
Find horizontal and vertical velocity after 2 s:
(25 m/s)
x ox
v v
2
0 ( 9.8 m/s )(2 s)
y oy
v v at
vx = 25.0 m/s
vy = -19.6 m/s
vx
vy
v0x = 25 m/s
v0y = 0
13. 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).
Note vertical and horizontal motions of balls
q
voy
vox
vo
vx = vox = constant
y oy
v v at
2
9.8 m/s
a
14. Displacement Calculations For
General Projection:
The components of displacement at time t are:
2
1
2
ox x
x v t a t
For projectiles: 0; ; 0;
x y oy ox o
a a g v v v
2
1
2
oy y
y v t a t
Thus, the displacement
components x and y for
projectiles are:
2
1
2
ox
oy
x v t
y v t gt
15. Velocity Calculations For
General Projection:
The components of velocity at time t are:
x ox x
v v a t
For projectiles: 0; ; 0;
x y oy ox o
a a g v v v
y oy y
v v a t
Thus, the velocity
components vx and vy
for projectiles are:
constant
x ox
y oy
v v
v v gt
16. Problem-Solving Strategy:
1. Resolve initial velocity vo into components:
vo
vox
voy
q cos ; sin
ox o oy o
v v v v
q q
2. Find components of final position and velocity:
2
1
2
ox
oy
x v t
y v t gt
Displacement: Velocity:
0
2
x x
y oy
v v
v v gt
17. Problem Strategy (Cont.):
3. The final position and velocity can be found
from the components.
R
x
y
q
4. Use correct signs - remember g is negative or
positive depending on your initial choice.
2 2
; tan
y
R x y
x
q
2 2
; tan
y
x y
x
v
v v v
v
q
vo
vox
voy
q
18. 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
30o
Since vx is constant, the horizontal displacements
after 2 and 4 seconds are:
(139 ft/s)(2 s)
ox
x v t
x = 277 ft
(139 ft/s)(4 s)
ox
x v t
x = 554 ft
0
(160 ft/s)cos30 139 ft/s
ox
v
0
(160 ft/s)sin30 80.0 ft/s
oy
v
19. 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: (Continued)
voy 160 ft/s
vox
30o
277 ft 554 ft
2 s 4 s
20. Example 2 (Cont.): Next we find the vertical
components of position after 2 s and after 4 s.
voy= 80 ft/s
160 ft/s
q
0 s 3 s
2 s
1 s 4 s
g = -32 ft/s2
y2
y4
2 2 2
1 1
2 2
(80 ft/s) ( 32 ft/s )
oy
y v t gt t t
The vertical displacement as function of time:
2
80 16
y t t
Observe consistent units.
21. (Cont.) Signs of y will indicate location of
displacement (above + or below – origin).
voy= 80 ft/s
160 ft/s
q
0 s 3 s
2 s
1 s 4 s
g = -32 ft/s2
y2
y4
Vertical position:
2
80 16
y t t
2
2 80(2 s) 16(2 s)
y 2
4 80(4 s) 16(4 s)
y
2 96 ft
y 4 16 ft
y
96 ft
16 ft
Each above origin (+)
22. (Cont.): Next we find horizontal and vertical
components of velocity after 2 and 4 s.
Since vx is constant, vx = 139 ft/s at all times.
Vertical velocity is same as if vertically projected:
2
; where g 32 ft/s
y oy
v v gt
At any
time t:
(32 ft/s)
y oy
v v t
139 ft/s
x
v
voy 160 ft/s
vox
30o
0
(160 ft/s)cos30 139 ft/s
ox
v
0
(160 ft/s)sin30 80.0 ft/s
oy
v
23. v2y = 16.0 ft/s
v4y = -48.0 ft/s
Example 2: (Continued)
vy= 80.0 ft/s
160 ft/s
q
0 s 3 s
2 s
1 s 4 s
g = -32 ft/s2
v2
v4
At any
time t:
(32 ft/s)
y oy
v v t
139 ft/s
x
v
80 ft/s (32 ft/s)(2 s)
y
v
80 ft/s (32 ft/s)(4 s)
y
v
24. At 2 s: v2x = 139 ft/s; v2y = + 16.0 ft/s
Example 2: (Continued)
vy= 80.0 ft/s
160 ft/s
q
0 s 3 s
2 s
1 s 4 s
g = -32 ft/s2
v2
v4
Moving Up
+16 ft/s
Moving down
-48 ft/s
The signs of vy indicate whether motion is
up (+) or down (-) at any time t.
At 4 s: v4x = 139 ft/s; v4y = - 48.0 ft/s
25. (Cont.): The displacement R2,q is found from
the x2 and y2 component displacements.
q
0 s 2 s 4 s
y2 = 96 ft
x2= 277 ft
R2
2 2
R x y
tan
y
x
q
2 2
(277 ft) (96 ft)
R
96 ft
tan
277 ft
q
R2 = 293 ft q2 = 19.10
t = 2 s
26. (Cont.): Similarly, displacement R4,q is found
from the x4 and y4 component displacements.
2 2
(554 ft) (64 ft)
R
64 ft
tan
554 ft
q
R4 = 558 ft q4 = 6.590
q
0 s 4 s
y4 = 64 ft
x4= 554 ft
R4
2 2
R x y
tan
y
x
q
t = 4 s
27. (Cont.): Now we find the velocity after
2 s from the components vx and vy.
2 2
2 (139 ft/s) (16 ft/s)
v
16 ft
tan
139 ft
q
v2 = 140 ft/s q2 = 6.560
voy= 80.0 ft/s
160 ft/s
q
0 s 2 s
g = -32 ft/s2
v2
Moving Up
+16 ft/s
v2x = 139 ft/s
v2y = + 16.0 ft/s
28. (Cont.) Next, we find the velocity after
4 s from the components v4x and v4y.
2 2
4 (139 ft/s) ( 46 ft/s)
v
16 ft
tan
139 ft
q
v4 = 146 ft/s q2 = 341.70
voy= 80.0 ft/s
160 ft/s
q
0 s 4 s
g = -32 ft/s2
v4
v4x = 139 ft/s
v4y = - 48.0 ft/s
29. Example 3: What are maximum height and
range of a projectile if vo = 28 m/s at 300?
ymax occurs when 14 – 9.8t = 0 or t = 1.43 s
Maximum y-coordinate occurs when vy = 0:
voy 28 m/s
vox
30o
ymax
vy = 0
2
14 m/s ( 9.8 m/s ) 0
y oy
v v gt t
vox = 24.2 m/s
voy = + 14 m/s
0
(28 m/s)cos30 24.2 m/s
ox
v
0
(28 m/s)sin30 14 m/s
oy
v
30. Example 3(Cont.): What is maximum height
of the projectile if v = 28 m/s at 300?
Maximum y-coordinate occurs when t = 1.43 s:
ymax= 10.0 m
voy 28 m/s
vox
30o
ymax
vy = 0
vox = 24.2 m/s
voy = + 14 m/s
2 2
1 1
2 2
14(1.43) ( 9.8)(1.43)
oy
y v t gt
20 m 10 m
y
31. Example 3(Cont.): Next, we find the range
of the projectile if v = 28 m/s at 300.
The range xr is defined as horizontal distance
coinciding with the time for vertical return.
voy 28 m/s
vox
30o
vox = 24.2 m/s
voy = + 14 m/s
Range xr
The time of flight is found by setting y = 0:
2
1
2 0
oy
y v t gt
(continued)
32. Example 3(Cont.): First we find the time of
flight tr, then the range xr.
voy 28 m/s
vox
30o
vox = 24.2 m/s
voy = + 14 m/s
Range xr
1
2 0;
oy
v gt
(Divide by t)
2
1
2 0
oy
y v t gt
xr = voxt = (24.2 m/s)(2.86 s); xr = 69.2 m
2
2(14 m/s)
;
-(-9.8 m/
2.86
s )
s
oy
t
v
t
g
33. 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
2 m
First find t from y equation:
0
½(-9.8)t2 = -(1.2)
t = 0.495 s
Note: x = voxt = 2 m
y = voyt + ½ayt2 = -1.2 m
2
1
2 1.2 m
y gt
2( 1.2)
9.8
t
R
34. Example 4 (Cont.): We now use horizontal
equation to find vox leaving the table top.
Use t = 0.495 s in x equation:
v = 4.04 m/s
1.2 m
2 m
R
Note: x = voxt = 2 m
y = ½gt2 = -1.2 m
2 m
ox
v t
2 m
(0.495 s) = 2 m;
0.495 s
ox ox
v v
The ball leaves the
table with a speed:
35. Example 4 (Cont.): What will be its speed
when it strikes the floor?
vy = 0 + (-9.8 m/s2)(0.495 s)
vy = vy + gt
0
vx = vox = 4.04 m/s
Note:
t = 0.495 s
vy = -4.85 m/s
2 2
(4.04 m/s) ( 4.85 m/s)
v
4.85 m
tan
4.04 m
q
v4 = 146 ft/s q2 = 309.80
1.2 m
2 m vx
vy
36. Example 5. Find the “hang time” for the football
whose initial velocity is 25 m/s, 600.
vo =25 m/s
600
y = 0; a = -9.8 m/s2
Time of
flight t
vox = vo cos q
voy = vo sin q
Initial vo:
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.
2 2
1 1
2 2
; 0 (21.7) ( 9.8)
oy
y v t at t t
37. vo =25 m/s
600
y = 0; a = -9.8 m/s2
Time of
flight t
vox = vo cos q
voy = vo sin q
Initial vo:
2 2
1 1
2 2
; 0 (21.7) ( 9.8)
oy
y v t at t t
4.9 t2 = 21.7 t 4.9 t = 21.7
2
21.7 m/s
4.9 m/s
t t = 4.42 s
Example 5 (Cont.) Find the “hang time” for the
football whose initial velocity is 25 m/s, 600.
38. Example 6. A running dog leaps with initial
velocity of 11 m/s at 300. What is the range?
v = 11 m/s
q =300
Draw figure and
find components:
vox = 9.53 m/s
voy = 5.50 m/s vox = 11 cos 300
voy = 11 sin 300
2 2
1 1
2 2
; 0 (5.50) ( 9.8)
oy
y v t at t t
To find range, first find t when y = 0; a = -9.8 m/s2
4.9 t2 = 5.50 t
2
5.50 m/s
4.9 m/s
t t = 1.12 s
4.9 t = 5.50
39. Example 6 (Cont.) A dog leaps with initial
velocity of 11 m/s at 300. What is the range?
v = 10 m/s
q =310
Range is found
from x-component:
vx = vox = 9.53 m/s
x = vxt; t = 1.12 s vox = 10 cos 310
voy = 10 sin 310
Horizontal velocity is constant: vx = 9.53 m/s
Range: x = 10.7 m
x = (9.53 m/s)(1.12 s) = 10.7 m
40. Summary for Projectiles:
1. Determine x and y components v0
cos and sin
ox o oy o
v v v v
q q
2. The horizontal and vertical components of
displacement at any time t are given by:
2
1
2
ox oy
x v t y v t gt
41. Summary (Continued):
4. Vector displacement or velocity can then
be found from the components if desired:
3. The horizontal and vertical components of
velocity at any time t are given by:
;
x ox y oy
v v v v gt
2 2
R x y
tan
y
x
q
