1. Review

2. Acceleration due to Gravity

3. Historical Perspective
Galileo was the first to show that all objects fall to the earth with the constant acceleration.
There is no evidence that he actually dropped bowling balls from the leaning tower of Piza.
He conducted experiments with balls rolling down inclined planes.

4. Objectives
Students will describe the motion of an object in free fall from rest in terms of distance fallen each second, speed, and acceleration
Students will measure the acceleration of a fallen object.
Students will solve for all variables in given equation dy= ½ gt2.
Students will express the rate of a falling objects quantitatively.

5. Gravitational Pull
When air resistance can be ignored then acceleration due to gravity is the same for all objects at the same location on earth.

6. B. Acceleration due to gravity is represented by the symbol g.

7. C. Since acceleration is a vector quantity, it must have both magnitude and direction.

8. D. We use up as our positive direction, so falling objects have a negative acceleration.

10. Equations vf = vi + gt vf 2 = vi2 + 2gdy dy= vit + ½ gt2
E. Assuming no air resistance, all falling objects in motion can be solved by replacing acceleration a with g for gravity.
vf = vi + gt
vf 2 = vi2 + 2gdy
dy= vit + ½ gt2

11. F. Free fall is defined as an object free to move under the influence of gravity, not that it is released from rest.

12. Earth’s gravitational Pull
g on earth = m/s2

13. Example Problem #1
The time the Demon Drop ride at Cedar Hall, Ohio is freely falling is 1.5 s.
A. What is its velocity at the end of this time?
B. How far does it fall?

14. Equation Part a Given: g= 9.8 m/s2
vi= 0 m/s
t= 1.5 s
Unknown: vf
vf = vi + gt
vf=0m/s+(-9.8m/s2x1.5s)
vf = -15 m/s
The time the Demon Drop ride at Cedar Hall, Ohio is freely falling is 1.5 s.
A. What is its velocity at the end of this time?

15. Equation Part b
The time the Demon Drop ride at Cedar Hall, Ohio is freely falling is 1.5 s.
B. How far does it fall?
Given:
g= 9.8 m/s2
vi= 0 m/s vf = -15 m/s
t= 1.5 s
Unknown: d
dy= vit + ½ gt2
dy=0m/s(1.5s)
+(1/2-(9.8m/s2)x(1.5s) 2)
dy= -11 m

16. Example Problem #2
King Kong carries Fay Wray up the 321m tall Empire State Building . At the top of the sky scrapper, Fay Wray’s shoe falls from her foot.
How fast will the shoe
be moving when it hits
the ground?

17. Example Problem #2
King Kong carries Fay Wray up the 321m tall Empire State Building . At the top of the sky scrapper, Fay Wray’s shoe falls from her foot.
How fast will the shoe be moving when it hits the ground?
Given:
g= 9.8 m/s2
vi= 0 m/s d= 321m
Unknown: vf
vf 2 = vi2 + 2gdy
vf 2 =0m/s 2 +2(9.8m/s2)x(321m)
vf = √
vf =

18. Example Problem # 3
The Steamboat Geyser in Yellowstone Park, Wyoming is capable of shooting its hot water up from the ground with a
speed of 48.0 m/s.
How high can
the geyser shoot?

19. Example # 3
The Steamboat Geyser in Yellowstone Park, Wyoming is capable of shooting its hot water up from the ground with a speed of 48.0 m/s. How high can the geyser shoot?
Given:
g= 9.8 m/s2
vi= 48.0 m/s vf =0 m/s
Unknown: d
dy= vf 2 -vi 2 /2g
dy= -115m

20. Example # 4
A baby blue jay sits in a tree awaiting the arrival of its dinner. As the mother lands on the nest, she drops a worm toward the hungry chick’s mouth, but the worm misses and falls from the nest to the ground in 1.5s.
How high up is the tree?

21. Example # 4 Given: t= 1.5s g= 9.8 m/s2
A baby blue jay sits in a tree awaiting the arrival of its dinner. As the mother lands on the nest, she drops a worm toward the hungry chick’s mouth, but the worm misses and falls from the nest to the ground in 1.5s. How high up is the tree
Given:
t= 1.5s
g= 9.8 m/s2

22. Example #5
A giraffe, who stands 6m tall, bites a branch off a tree to chew on the leaves, and he lets the branch fall to the ground. How long does it take the branch to hit the ground?

23. Example #5
A giraffe, who stands 6m tall, bites a branch off a tree to chew on the leaves, and he lets the branch fall to the ground. How long does it take the branch to hit the ground?
Given:
dy=6m
g=9.8 m/s2
vo= 0 m/s
Unknown t=?
Equation

24. Think about it
If a penny falls from the top of the Empire State Building will it injure a passerby on the street below?

31. Graphing ! x t A B C
1 – D Motion
A … Starts at home (origin) and goes forward slowly
B … Not moving (position remains constant as time progresses)
C … Turns around and goes in the other direction quickly, passing up home

42. Real life
Note how the v graph is pointy and the a graph skips. In real life, the blue points would be smooth curves and the green segments would be connected. In our class, however, we’ll mainly deal with constant acceleration. v t a t

43. Area under a velocity graph
“forward area”
“backward area”
Area above the time axis = forward (positive) displacement.
Area below the time axis = backward (negative) displacement.
Net area (above - below) = net displacement.
Total area (above + below) = total distance traveled.

46. v (m/s)
Area units 12
t (s)
Imagine approximating the area under the curve with very thin rectangles.
Each has area of height  width.
The height is in m/s; width is in seconds.
Therefore, area is in meters!
12 m/s
0.5 s
The rectangles under the time axis have negative heights, corresponding to negative displacement.

47. Graphs of a ball thrown straight upx
The ball is thrown from the ground, and it lands on a ledge.
The position graph is parabolic.
The ball peaks at the parabola’s vertex.
The v graph has a slope of -9.8 m/s2.
Map out the slopes!
There is more “positive area” than negative on the v graph. t v t a t

48. Graph Practice Try making all three graphs for the following scenario:
1. Schmedrick starts out north of home. At time zero he’s driving a cement mixer south very fast at a constant speed.
2. He accidentally runs over an innocent moose crossing the road, so he slows to a stop to check on the poor moose.
3. He pauses for a while until he determines the moose is squashed flat and deader than a doornail.
4. Fleeing the scene of the crime, Schmedrick takes off again in the same direction, speeding up quickly.
5. When his conscience gets the better of him, he slows, turns around, and returns to the crash site.

49. Kinematics Practice
A catcher catches a 90 mph fast ball. His glove compresses 4.5 cm. How long does it take to come to a complete stop? Be mindful of your units!
2.24 ms
Answer

50. Uniform Acceleration
x = 1
x = 3
x = 5
x = 7
t :
x :
( arbitrary units )
When object starts from rest and undergoes constant acceleration:
Position is proportional to the square of time.
Position changes result in the sequence of odd numbers.
Falling bodies exhibit this type of motion (since g is constant).

51. Spreadsheet Problem
We’re analyzing position as a function of time, initial velocity, and constant acceleration.
x, x, and the ratio depend on t, v0, and a.
x is how much position changes each second.
The ratio (1, 3, 5, 7) is the ratio of the x’s.
Make a spreadsheet like this and determine what must be true about v0 and/or a in order to get this ratio of odd numbers.
Explain your answer mathematically.

52. Let’s use the kinematics equations to answer these:
Relationships
Let’s use the kinematics equations to answer these:
1. A mango is dropped from a height h.
a. If dropped from a height of 2 h, would the impact speed double?
Would the air time double when dropped from a height of 2 h ?
A mango is thrown down at a speed v.
If thrown down at 2 v from the same height, would the impact speed double?
Would the air time double in this case?

53. Relationships (cont.)
A rubber chicken is launched straight up at speed v from ground level. Find each of the following if the launch speed is tripled (in terms of any constants and v).
max height
hang time
impact speed
9 v2 / 2 g
6 v / g
3 v
Answers