1. Chapter 3 Kinematics in 2D
1. SHOOT AND DROP BOTH VERSIONS: DROP A BALL AND SHOOT A BALL OUT STRAIGHT. LAND AT THE SAME TIME
2. PROJECTILE MOTION. TWO BALLS ARE RELEASED ON A WOODEN TRACK AND ONE FALLS ON THE OTHER
3. MONKEY AND THE GUN USING A DART GUN.
4, REL;ATIVITY CAR
x and y components

2. Math Review First Vectors and Unit Vectors
Representation of a vector : has magnitude and direction
i and j are unit vectors
x and y are components
angle gives direction and length of vector gives the magnitude
Example of vectors
Addition and subtraction
Scalar or dot product

3. Vectors Red arrows are the i and j unit vectors. Magnitude =
The angle between A and the x axis is given by

4. Adding Two Vectors Create a Parallelogram with The two vectors
You wish you add.

5. Adding Two Vectors
Note you add x and
y components .

6. Vector components in terms of sine and cosiney x θ r

7. Scalar product =
90 deg θ AB
Also

8. AB is the perpendicular projection of A on B. Important later.θ AB
Also

9. Vectors in 3 Dimensions

10. For a Right Handed 3D-Coordinate Systemsj k x z
Right handed rule.
Also called cross product
Magnitude of

11. Suppose we have two vectors in 3D and we want to add themy i j k r1 r2 2 -3 4 1 5 x 7 z

12. Adding vectors
Now add all 3 components i j k x y z r r2 r1

13. Scalar product =
The dot product is important in the of discussion of work.
Work = Scalar product

14. Cross Product =
See your textbook for more information on vectors and cross and dot products

15. Now we want to use what we know about x and y components of a vector to help understand motion in the xy plane: We call this Projectile Motion
Galileo separated motion into Horizontal and Vertical
Horizontal: v is constant because a = 0
Vertical: a is constant because a = g
(assume air resistance or friction is 0)
Demo of projectile balls

16. x direction
3 independent equations
Derive these 2 from the other 3

17. y direction
3 independent equations
Derive these 2 from the other 3

18. Consider a ball projected horizontally with initial speed v0 ax= 0 and ay= -g. Initial position is (x0,y0) . y v0 x0
The equations that describe the x and y components of the motion of the upper ball in the previous experiment are familiar to us. In the x direction the velocity is constant, so x increases in proportion to t. In the y direction we have a falling body dropped from a height y0. We can eliminate t from the two equations by solving the first one for t and substituting it in the second. The result is a parabola.
The trajectory of a body moving with a constant horizontal velocity and constant vertical acceleration is a parabola. This is referred to as projectile motion. Examples are: baseball, football, tennis, shooting a bullet, etc. In our treatment we are ignoring the force of the air on the moving body. As we have seen, this is only accurate when the speed of motion is small enough. y0 x
projectile ball

19. Demo Problem
A projectile at y0=1.5 m high is shot horizontally with speed v0 that causes it to land 1.25 m away from the nozzle. What is the time in the air? y v0 x0 y0
1.5 m
t = time in the air:
Here is a simple example of projectile motion. The rock thrown from the building continues moving with the same horizontal speed throughout its motion. All we need to know is how long was it in the air?
This can be found using its vertical motion which is the same as if it had just been dropped. From this we find that it lands after 2 seconds. And now we know how far it moved horizontally during this time. x
1.25 m

20. Demo Problem
A projectile at 1.5 m high is shot horizontally with speed v0 that causes it to land 1.25 m away from the nozzle.
What is v0? y v0 x0 y0
1.5 m
Here is a simple example of projectile motion. The rock thrown from the building continues moving with the same horizontal speed throughout its motion. All we need to know is how long was it in the air?
This can be found using its vertical motion which is the same as if it had just been dropped. From this we find that it lands after 2 seconds. And now we know how far it moved horizontally during this time. x
1.25 m

21. Example What is the speed v when it lands? y v0 x0 1.5 m y0 x 1.25 m
Here is a simple example of projectile motion. The rock thrown from the building continues moving with the same horizontal speed throughout its motion. All we need to know is how long was it in the air?
This can be found using its vertical motion which is the same as if it had just been dropped. From this we find that it lands after 2 seconds. And now we know how far it moved horizontally during this time. x
1.25 m

22. Demo
Find vy ? y v0 x0
1.5 m y0
Here is a simple example of projectile motion. The rock thrown from the building continues moving with the same horizontal speed throughout its motion. All we need to know is how long was it in the air?
This can be found using its vertical motion which is the same as if it had just been dropped. From this we find that it lands after 2 seconds. And now we know how far it moved horizontally during this time. x
1.25 m v
What is the impact angle when the ball strikes the ground?

23. Projectile Motion y v0
v0y θ0
v0x x

24. Projectile Motion y v0
v0y θ0
v0x x

26. Projectile Motion

27. Rules for doing projectile motion problems
Draw a careful diagram
Choose origin and xy coordinate system
Examine x and y motion - maybe you want to resolve
motion into components.
4. List known and unknown quantities.
Usually ax = 0 and ay = - g if y positive is up
vx = constant everywhere and
vy=0 at highest point of trajectory
5. Apply the relevant equations. Combine components
into a vector to get final answer sometimes

28. a) How far does it go when it reaches the batter?
Slow Pitch Problem
0.25 s R
A baseball is released at 3 ft above ground level. A stroboscopic plot shows
the position of the ball every 0.25 sec. Answer the following questions.
a) How far does it go when it reaches the batter?
b) How long does it take the ball to reach the bat?
c) What is the initial speed of the ball?

29. c) What is the initial speed of the ball?R θ0 θ0
c) What is the initial speed of the ball?
To find v0y. Consider what happens when the ball reaches the batter.
v0 = (v20x+v20y)1/2
Find v0x v0
v0y
v0x θ0
FINALLY

30. R H
d) How high does the ball go above the ground? It reaches the maximum when vy = 0.
H = (36 )2(0.44)2/2(32) =
(Above the ground)

31. e) What is the launch angle?
Slow Pitch Problem
0.25 s R H θ0
e) What is the launch angle? v0
v0y
v0x θ0

32. Foul Shot problem Find the initial speed and the time it takes
for the ball to go through the basket.

33. Foul Shot problem Find the initial speed and the time it takes
for the ball to go through the basket.

34. Foul Shot Continued Find the initial speed and
find the time it takes for the ball to go through the basket

35. Horizontal Range formula
A cannon ball is fired with speed v0 = 82 m/s at a ship 560 m from shore.
What are the two launch angles needed to hit the ship?

36. Horizontal Range formula
A cannon ball is fired with speed v0 = 82 m/s at a ship 560 m from shore.
What are the two launch angles needed to hit the ship?
x component
y component
Solve for t

37. Horizontal Range continued
θ0 = 27.3 degrees
second solution 2θ0 = = degrees
θ0 = 62.6 degrees

38. Three Important Points
Inertial reference frame - Where Newton’s Laws hold true.
One that is not accelerating- not the earth.
Acceleration is the same in any two frames that are moving at constant velocity relative to each other.
Principle of superposition

39. Two Reference frames moving at a Relative Speed VW/E with respect to each other
Speeds are always measured with respect to an origin of coordinates, or a reference frame. Sometimes that reference frame might itself be moving with respect to something else. How do we take a moving reference frame into account? The Galilean method is illustrated here. A wagon moves along the ground with a speed vWE, the speed of the wagon with respect to the Earth. A man walks on the wagon with a speed vMW, the speed of the man with respect to the wagon. What is the man’s speed with respect to the Earth? It is simply the sum of the above two speeds.
The man could of course be walking backwards along the wagon, in which case vMW would be negative. Labeling the speeds this way provides a mnemonic to help remember the result, and to help keep signs straight. The moving reference frame labeled by W in this case, appears twice on the right side of the equation, sandwiched between M and E which appear on the left side. The first and last subscripts on the left side are the same as the first and last on the right. E
Frame W is attached to the wagon
Frame E is attached to the Earth
Constant VW/E
aW/E = 0

40. An observer in frame W measures the speed VM/W relative to the wagon
VW/E
Speeds are always measured with respect to an origin of coordinates, or a reference frame. Sometimes that reference frame might itself be moving with respect to something else. How do we take a moving reference frame into account? The Galilean method is illustrated here. A wagon moves along the ground with a speed vWE, the speed of the wagon with respect to the Earth. A man walks on the wagon with a speed vMW, the speed of the man with respect to the wagon. What is the man’s speed with respect to the Earth? It is simply the sum of the above two speeds.
The man could of course be walking backwards along the wagon, in which case vMW would be negative. Labeling the speeds this way provides a mnemonic to help remember the result, and to help keep signs straight. The moving reference frame labeled by W in this case, appears twice on the right side of the equation, sandwiched between M and E which appear on the left side. The first and last subscripts on the left side are the same as the first and last on the right. E
An observer in frame E measures the speed VM/E relative to the Earth and both are related by VW/E
VM/E = VM/W + VW/E

41. Acceleration and forces for an observer on Earth are the same
VW/E
VM/W W E
Constant VW/E
aW/E = 0
VM/E = VM/W + VW/E
aM/E = aM/W
Speeds are always measured with respect to an origin of coordinates, or a reference frame. Sometimes that reference frame might itself be moving with respect to something else. How do we take a moving reference frame into account? The Galilean method is illustrated here. A wagon moves along the ground with a speed vWE, the speed of the wagon with respect to the Earth. A man walks on the wagon with a speed vMW, the speed of the man with respect to the wagon. What is the man’s speed with respect to the Earth? It is simply the sum of the above two speeds.
The man could of course be walking backwards along the wagon, in which case vMW would be negative. Labeling the speeds this way provides a mnemonic to help remember the result, and to help keep signs straight. The moving reference frame labeled by W in this case, appears twice on the right side of the equation, sandwiched between M and E which appear on the left side. The first and last subscripts on the left side are the same as the first and last on the right.
Acceleration and forces for an observer on Earth are the same
for an observer on the wagon as long as one frame
moves at constant speed relative to the other.

42. GALILEAN RELATIVITY
Person on a moving ship drops a rock. They see it fall straight
down to land at their feet.
Someone on shore sees the rock continue to move horizontally
as it falls, and says the trajectory is parabolic.
Another observer on another ship sees it move horizontally with
a different speed and sees a different parabola.
Galileo thought about the question of a moving Earth. If the Earth moves, why do we not notice this motion, as Aristotle claimed we would. Here is the thought experiment he carried out in answer to this puzzle.
A person on a ship drops a rock. To that person its trajectory is a straight line. An observer on shore sees the trajectory as a parabola. Someone on another ship sees yet a different parabola. All of them are making valid observations. They all see it land at the same time.
Galileo concluded that the vertical motion of the rock must in principle be independent of its horizontal motion. Furthermore he concluded that everything that happens on board the ship must be independent of its motion. He argued that whatever you take with you on the ship, insects, fish, physics experiments, etc will be unable to detect the horizontal motion of the ship. He was assuming, of course, that this motion is perfectly uniform and smooth.
This is called the Galilean principle of relativity and explains immediately why we do not sense the motion of the Earth about the Sun.
But they all see it land at the same time. Galileo concluded that the
horizontal motion cannot influence the vertical motion, and that
what takes place on the ship is independent of the ship’s motion.
He argued this applies to all physical and biological processes.
The same principle applies to what takes place on a moving Earth.

43. Example is the RELATIVITY CAR
DEMO RELATIVITY CART
This car with spring cannon illustrates Galileo’s principle of relativity. The cannon shoots a ball straight up. As it returns it is caught by the funnel.
When the car is stationary in the room, we see the ball go straight up and down. When the car moves relative to us, we see a parabolic trajectory because now the ball has a constant horizontal velocity along with its vertical free-fall motion. But the ball is still caught by the funnel because the ball and car have the same horizontal velocity. The car and ball move along together and the result is the same as if it were stationary in the room. Its uniform horizontal motion is unimportant, even undetectable by any experiment carried out on the cart.