1. Two-Dimensional Motion
Vertical & Horizontal

2. Vertical Motion
Galileo was the first to show that all objects fall to Earth with a constant acceleration.
Acceleration due to gravity (g) is the change in velocity over a period of time in a downward vertical direction.
Value of “g” = 9.80 m/s2

3. Directions and Signs A. Up v, t, & d are (+) g is (-)
B. Down v, d, & g are (-) t is (+)
“g” is negative regardless of direction.
“t” is always positive

4. Equations
V2 = V1 + gt
V22 = V gd
d = V1t + ½ (gt2)

5. Free-Fall I
An object dropped from rest takes 4 s to reach the ground. What is the final velocity of the object?
g(-) d(-) v(-) t (+) v1 = 0 m/s
v2 = v1 + gt
0 m/s + (-9.8 m/s2) (4s)
v2 = - 40 m/s
NO!

6. Free-Fall II
An object is thrown straight upward with a velocity of 10 m/s. What is the maximum height attained?
V2 = 0 m/s
V1 = +10 m/s
g = m/s2
V22 = V12 + 2gd
wee

7. Solution: Free –Fall II
d = V22 – V12
2 g
= (0 m/s)2 - (10 m/s)2
2 (-9.80 m/s2)
d = 5 m

8. Symmetry of Fall Time up = time down Distance up = distance down
Velocity up = velocity down

9. Testing Your Reaction Time
Ask a friend to hold a ruler just even with the top of your fingers.
Then have your friend drop the ruler.
Taking the number of centimeters the ruler falls before you can catch it, calculate your reaction time. Average three trials.

10. Reaction Time Results
The reaction time for most people is more than 0.15 s.

11. Blue Physics Textbook Read pages 76 – 79
Solve the problems listed below in your Classwork/Homework Notebook
p. 77: 25
p. 79:

12. Two-Dimensional Motion II
Object rolled horizontally
Components
Vx = value of vector
VY = 0 m/s
Displacement is a vector quantity.
d = Vxt Vx = horizontal velocity

13. Horizontal Velocity & Displacement
If the object is moving with constant velocity, the acceleration (horizontally) is zero.
velocity acceleration
Vx at beginning is equal to Vx at any point.

14. Path is called a trajectory
Vx = 10 m/s at all points along the
path
dx = Vxt

15. Vertical Velocity
Vertical velocity increases with time until the object reaches terminal velocity.
Terminal velocity - Velocity is no longer changing.

16. Vertical Displacement
Vertical displacement increases with time.
If d = V1t + ½(gt2) and V1 = 0 then
d = 1/2 gt2 = 4.9 m/s2 (t2)

17. Time (s)
Displacement (m) 1
4.9 2
19.6 3
44.1

18. Graphs of Motion Displacement is increasing with time.
Velocity is changing with time.
Object is accelerating in the “y” direction

19. Horizontal & vertical motions are independent of each other
Each type has its own equations
Both the X- and Y- components must be used to determine d2 and v2 at any point.**
Time down is equal to time across at all
points.
**Back to the RIGHT TRIANGLE

20. EQUATIONS X dX = vXt Y Use these equations for any point
on the trajectory
d = v1t + 1/2gt2
v2 = v1 + gt
v22 = v12 + 2gd
v2 =√ vx2 + vy2
tan Θ = vy/vx

21. EXAMPLE
A car runs off a 43.9 m cliff and lands 87.7 m from the base of the cliff.
a) How long does it take the car
to reach the ground?
What was the initial velocity of
the car just before it left the cliff?

22. t 2= (-43.9 m)
-9.80 m/s2
t = 2.99 s
d x= vxt vx = d/t = 87.7 m 2.99 s vx = 29.3 m/s

23. Velocity Upon Impact With Ground
Vx = m/s
Vy2 = vy1 + gt
= 0 m/s + (-9.80 m/s2)(2.99 s)
Vy = 29.3 m/s
Vf2 = vx2 + vy2
= (29.3 m/s)2 + (29.3 m/s)2
Vf = 41.4 m/s

24. Object Projected at an Angle
An object is shot into the air at a speed of 20. m/s and an angle of 40.o .

25. Things You Know Time up = time down
Acceleration in the vertical direction is always 9.80 m/s2.
Acceleration in the horizontal direction is always 0m/s2.
Velocity at the top of the flight is 0 m/s.

26. Velocity vector is acting at an angle
Use cos to determine the value of the
Vx component and sin for the Vy
component.
Vx doesn’t change throughout trip.
Velocity at release point = velocity at endpoint.

27. QUESTIONS AND SOLUTIONS
Step I: Calculate the x- and y-
components
Vx = V cos Θ = 20 m/s cos 40o
Answer: m/s
Vy = V sin Θ = 20 m/s sin 40o
Answer: 13 m/s

28. Step II: Calculate the time in air
t = vy2 – vy1 g
= -13 m/s – (+13 m/s)
-9.80 m/s2
t = 2.7 s

29. Calculate the Range (where the object will land)
d = Vxt
= (15 m/s)(2.7 s)
d = 41 m

30. Where is the object 0.5 s after being launched?
Step I: Calculate d in the y-direction
d = V1t + ½ at2
= 13 m/s (0.5s) + ½(9.8m/s2)(0.5s)2
dy = 5.3 m

31. Step II: Calculate d in the x-direction
dx = vx t
= (15 m/s) (0.5s)
dx = 7.5 m

32. Step III: Calculate the Resultant Displacement
df2 = dx2 + dy2
= (7.5 m) (5.3 m)2
df = 9.2 m

33. Calculate the Angle
tan Θ = dy/dx
= 5.3 m ÷ 7.5 m
= 0.688
= 35o