1. Motion in Two Dimensions
Holt Physics
Pages 98 – 106

2. Things to Remember
one-dimensional motion was either all in the horizontal (x) direction or all in the vertical (y) direction
acceleration of gravity (ag or g) is -9.8 m/s2 in the downward direction
gravity acts the SAME on ALL objects!!
downward velocity or position represented by “-“
find the components of any vector the same way as previous section

3. Also remember…

4. Equations Recall from chapter 2 when we studied 1-dimensional motion:
v = Δx/Δt (used with constant velocity)
Δy = vΔt + ½ aΔt2
vf2 = vi2 + 2aΔy
c2 = a2 + b2

5. Reviewing the Math #1
If you drop your cell phone from the top of your desk as you try to hide it (0.82 m), how long will it take to hit the ground?

6. Reviewing the Math #2
Find the two components of the velocity for each vector: (vx and vy)
v = 25 m/s horizontally
v = 25 m/s at 20° N of E

7. Motion of Objects Projected Horizontally
Chapter (Part 1)
Motion of Objects Projected Horizontally

8. Objective 1 Understand how to separate a motion vector into its horizontal and vertical components Objective 2 Understanding that motions which are perpendicular to each other are also independent of each other Objective 3 Solve problems involving objects projected horizontally

9. Projectiles Projectile: Projectile Motion:
objects thrown or launched into the air and subject to gravity
Projectile Motion:
motion through the air without a propulsion

10. Projectiles Curved Motion: Resolution into Components:
path is parabolic (larger curve for higher launches)
Resolution into Components:
projectile motion can be separated into both vertical and horizontal components
Initial vertical component = viy = 0m/s
(object in free-fall, a = g = -9.8m/s2)
Initial horizontal component = vix = vx = shot at…

11. Components of a Horizontal Projectile
vx (horizontal velocity) is constant and there is no acceleration in this direction (we ignore air resistance)
vy is not constant, it is accelerated downward due to gravity (9.8 m/s2 down or -9.8 m/s2, where the - sign means down)

12. This means:
In the horizontal direction, equal distances are covered in equal amounts of time
In the vertical direction, there is acceleration, so different distances are covered in equal amounts of time

13. y v0 x

14. y x

15. y x

16. y x

17. y x

18. g = -9.81m/s2 y Object is in free-fall
Acceleration is constant, and downward ay= g = -9.81m/s2
The horizontal (x) component of velocity is constant
The horizontal and vertical motions are independent of each other, but they have a common time
g = -9.81m/s2 x

19. WHAT DOES THIS MEAN? Remember: vy = aΔt
What are the components at each second?
t = 0s s s s 4s

20. Horizontal and vertical velocities are completely independent of one another!

21. ax = 0 ay = g = -9.81 m/s2 vi = vx vy = ayΔt vyf2 = 2ayΔy Δx = vxΔt
Equations of motion:
horizontal (x)
uniform motion
vertical (y)
accelerated motion
acceleration
ax = 0
ay = g = m/s2
velocity
vi = vx
vy = ayΔt
vyf2 = 2ayΔy
displacement
Δx = vxΔt
Δy = ½ ayΔt2
Frame of reference: x y vi h or Δy
g or a

24. Analysis of Motion ASSUMPTIONS:
x-direction (horizontal): uniform motion
y-direction (vertical): accelerated motion
no air resistance
POSSIBLE QUESTIONS:
What is the total time of the motion?
What is the horizontal range (Δx)?
What is the initial velocity?
What is the final velocity?

25. How long does it take the football to reach the bottom of the cliff?
A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high
How long does it take the football to reach the bottom of the cliff?
vx= 5m/s
78.4 m

26. How long does it take the football to reach the bottom of the cliff?
A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high
How long does it take the football to reach the bottom of the cliff?
Unknown: Δt
Given:
yf = 78.4m
yi = 0 m
vyi = 0 m/s
vxi = 5.0 m/s
a = -9.8m/s2
vx= 5m/s
Equation:
Δy = ½ aΔt2
Solving for Δt
Δt2 = 2 Δyf/a
Δt = √ 2Δyf/a
Δt = √ 2(-78.4m)/(-9.8m/s2)
Δt = 4.0 s
78.4 m

27. How far from the base of the cliff does the football land?
A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high
How far from the base of the cliff does the football land?
vx= 5m/s
78.4m ?

28. How far from the base of the cliff does the football land?
A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high
How far from the base of the cliff does the football land?
Unknown: Δx
Given:
yf = 78.4m
yi = 0 m
vyi = 0 m/s
vxi = 5.0 m/s
a = -9.8 m/s2
Δt = 4.0 s
vx= 5m/s
Equation:
Δ x = vx Δt
78.4m
Solving for Δ x
Δ x = (5.0m/s)(4.00s)
Δ x = 20. m ?

29. A football is thrown horizontally at a speed of 5
A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high
What are the horizontal and vertical components of the velocity just before it hits the ground (4sec)?
vx= 5m/s
78.4m vf

30. A football is thrown horizontally at a speed of 5
A football is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high
What are the horizontal and vertical components of the velocity just before it hits the ground (4sec)?
Unknown: vfx , vfy
Given:
vxi =5.0 m/s yi = 0 m
yf = 78.4m
vyi = 0 m/s
Δt = 4.0 s
a = -9.8 m/s2
vx= 5m/s
Equation for vxf
vxf = vxi
Solving
vxf = 5.0 m/s
Equation for vyf
vyf = a Δt
Solving
vyf = (-9.8 m/s2 )(4s)
vyf = -39 m/s
78.4m vf

31. Sample Problem
Someone is being chased down a river by someone else in a faster craft. Just as the fast boat pulls up to the slower boat, both reach the edge of a 5.0 m waterfall. If the slower boat’s velocity is 15 m/s and the faster boat’s speed is 26 m/s, how far apart will the two vessels be when they land?

32. sketch known(x) known(y) unknown
equation/solution
known(x) known(y) unknown
equation/solution

33. Sample Problem
An African Spitting Cobra can raise its head straight up approximately 0.61m. An average distance that the poisonous spit travels is 3.60m. What is the horizontal velocity of this deadly venom?
sketch known(x) known(y) unknown
equation/solution

34. Sample Problem
A girl jumps off of the 10. m platform with a horizontal velocity of 2.0 m/s. How far from the end of the platform does she hit the water?

35. (Δt = 1.43 sec)

36. Sample Problem
An army helicopter needs to drop supplies to troops in the field. If the army helicopter is flying at an altitude of 500 m and a horizontal velocity of 10 m/s, how far before it gets to the target zone should they drop the supplies?

37. (Δx = 101m)