1. Two-Dimensional Motion Projectiles launched at an angle
Chapter 3
Two-Dimensional Motion
Projectiles launched at an angle

2. Some Variations of Projectile Motion
An object may be fired horizontally
The initial velocity is all in the x-direction
vi = vx and vy = 0
All the general rules of projectile motion apply

3. Projectile Motion at an angle

4. How are they different? Projectiles Launched Horizontally
The initial vertical velocity is 0.
The initial horizontal velocity is the total initial velocity.
Projectiles Launched At An Angle
Resolve the initial velocity into x and y components.
The initial vertical velocity is the y component.
The initial horizontal velocity is the x component.

5. Some Details About the Rules
x-direction
ax = 0
vx = vix = vicosΘi = constant
x = vixt
This is the only equation in the x-direction since there is constant velocity in that direction
Initial velocity still equals final velocity

6. More Details About the Rules
y-direction
viy = visinΘi
Free fall problem
a = g
Object slows as it goes up (-9.8m/s2)
Uniformly accelerated motion, so the motion equations all hold
JUST LIKE STOMP ROCKETS
Symmetrical

7. Problem-Solving Strategy
Resolve the initial velocity into x- and y-components
Treat the horizontal and vertical motions independently
Make a chart again, showing horizontal and vertical motion
Choose to investigate up or down.
Follow rules of kinematics equations

8. Solving Launched Projectile Motion
vi = (r , Θ) = (vix , viy)
Horizontal Vertical
a =
vi =
vf =
t =
x =
vix #
# = range
a =
vi =
vf =
t =
x =
+/- 9.8m/s2 = g
0 or viy
viy or 0 #
Max height = y
UP or DOWN INVESTIGATION… Where do the resolved components go?

9. Projectile Motion at an angle

10. Example 1:
The punter for the Steelers punts the football with a velocity of 27 m/s at an angle of 30. Find the ball’s hang time, maximum height, and distance traveled (range) when it hits the ground. (Assume the ball is kicked from ground level.)
Looking for:
Total time (t)
Max height (y)
Range (x)
Given:
vi = (27m/s, 30o)

11. What do we do with the given info?
vi = (27m/s, 30o)
What are the units?
vi = (23.4m/s, 13.5m/s)
“resolved” vector
m/s
27m/s
30o
Viy = 27sin30
Viy = 13.5
Viy = 13.5m/s
Vix = 27cos30
Vix = 23.4
Vix = 23.4m/s

12. So where does this info “fit” in the chart?
Horizontal Vertical
a =
vi =
vf =
t =
x =
23.4m/s
a =
vi =
vf =
t =
x =
Viy if solving “up” = 13.5m/s
Viy if solving “down” = 13.5m/s

13. Pick a “side” to solve – symmetry
Up:
Horizontal Vertical
a =
vi =
vf =
t =
x =
23.4m/s
a =
vi =
vf =
t =
y =
- 9.8m/s2
13.5m/s
vf2 = vi2 + 2gy
y = 9.3m
vf = vi + gt
t = 1.38s

14. Projectile Motion at an angle

15. Pick a “side” to solve – symmetry
Down:
Horizontal Vertical
a =
vi =
vf =
t =
x =
23.4m/s
a =
vi =
vf =
t =
y =
9.8m/s2
13.5m/s
vf2 = vi2 + 2gy
y = 9.3m
vf = vi + gt
t = 1.38s

16. Projectile Motion at an angle

17. On the horizontal side Max height occurs midway through the flight.
We found t = 1.38s both directions (up and down).
How long is the projectile in the air? DOUBLE this time for total air time t = 1.38 x 2 = 2.76s
What about range? x = vt x = (23.4)(2.76) x = 64.6m = RANGE

18. This tells us…
Now we only need an initial velocity vector to determine all of the information we need to have a detailed description of where an object is in its path.
vi = (r , Θ) = (vix , viy)

19. Maximum Range vs. Maximum Height
What angle of a launched projectile gets the maximum height?
What angle of a launched projectile gets the maximum range?
90o
45o

20. Projectile Motion at Various Initial Angles
Complementary values of the initial angle result in the same range
The heights will be different
The maximum range occurs at a projection angle of 45o

21. Non-Symmetrical Projectile Motion
Follow the general rules for projectile motion
Break the y-direction into parts
up and down
symmetrical back to initial height and then the rest of the height

22. your homework …
We are going to see what kind of job Hollywood writers and producers would do on their NECAP assessments…
Watching a clip of the Bus Jump, use the timer provided to time the flight of the bus and then do the actual calculations for homework.