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### Pointing the Way

### Checking Your References

### Fire Away!!!

### Over the Edge

Do now – on a new piece of paper

- The diagrams below represent two types motions. One is constant motion, the other, accelerated motion. Which one is constant motion and which one is accelerated motion? explain your answer.

Objectives

1. Distinguish between a scalar and a vector.

2. Add and subtract vectors using the graphical method.

3. Multiply and divide vectors by scalars.

Homework – vector 1

Vectors

Representing Vectors

Vectors on paper are simply arrows

Direction represented by the way the ARROW POINTS

Magnitude represented by the ARROW LENGTH

Examples of Vectors

Displacement

Velocity

Acceleration

Vector Diagrams

- a scale is clearly listed
- a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a head and a tail.
- the magnitude of the vector is clearly labeled.

head

tail

Vectors can be moved parallel to themselves in a diagram

Directions of Vector

Reference Vector

Uses due EAST as the 0 degree reference,

all other angles are measured from that point

20 meters at 190°

34 meters at 48°

Compass Point

- The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from east, west, north, or south

20 meters at 10° south of west

34 meters at 42° east of north

90°

N

180°

W

E

0°

S

270°

Changing Systems

What is the reference vector angle for a vector that points 50 degrees east of south?

What is the reference vector angle for a vector that points 20 degrees north of east?

50°

20°

270° + 50° = 320°

20°

Practice

- Guided notes – page 2 – 1a, 2a

What we can DO with vectors

ADD/SUBTRACTwith a vector

To produce a NEW VECTOR

MULTIPLY/DIVIDEby a vector or a scalar

To produce a NEW VECTOR or SCALAR

Vector Addition

- Two vectors can be added together to determine the sum (or resultant).

The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together

+

= ?

A

B

A

B

Two methods for adding vectors- Graphical method: using a scaled vector diagram
- The head-to-tail method
- Parallelogram method
- Mathematical method - Pythagorean theorem and trigonometric methods

A

B

C

A

B

C

Vector addition: head-to-tail method- A cart is pushed in two directions, as the result, the cart will move in the resultant direction

+

=

+

=

(Resultant)

A

B

A

B

30o

60o

- Frank is walking from his house to Adam’s house, which 200 m away at 30o north of east from his house. Frank and Adam then walk together to their school. The school is 200 m away at 60o west of north from Adam’s house. Determine the resultant (vector sum) of the two vectors ( : Frank’s house to Adam's house; : Adam's house to school)

school

The resultant is not from the head to tail, it is from beginning to end.

Adam’s house

resultant

Frank’s house

Magnitude: measure with ruler, determine using scale

Direction: measure with protractor with East

A

B

C

Parallelogram method

- A cart is pushed in two directions, as the result, the cart will move in the resultant direction

10/5 do now Determine resultant graphically

- Vector A: 3 m East; Vector B: 6 m North
- You need to
- Indicate your scale
- show magnitude and direction of the resultant

objectives

- Lab 6 – motion graphs
- Vector addition – graphically and mathematically
- Homework
- Castle learning
- Vector HW 2

Lab 6 – graph matching

- OBJECTIVES
- Analyze the motion of a student walking across the room.
- Predict, sketch, and test position vs. time kinematics graphs.
- Predict, sketch, and test velocity vs. time kinematics graphs.
- MATERIALS
- Computer; Motion Detector; Vernier computer interface; Logger Pro

PRELIMINARY QUESTIONS

- An object at rest
- An object moving in the positive direction with a constant speed
- An object moving in the negative direction with a constant speed
- An object that is accelerating in the positive direction, starting from rest

match the graphs with the following motion

position

position

position

position

velocity

velocity

velocity

velocity

Time

Time

Time

Time

Time

Position vs. Time and Velocity vs. Time Graph Matching

- Sketch a Position vs. Time graph and Velocity vs. Time graph that that represents the motion of the object:
- Speeds up from rest to the positive direction; Moves at constant speed to positive direction; Slows down to a stop; Speeds up from rest to the negative direction; Moves at constant speed to the negative direction; Slows down to a stop.

Compare your sketch with the experimental data. If you had made mistakes, sketch the correct graph and explain your mistakes.

ANALYSISPART I – POSITION VS. TIME GRAPH

- Explain the significance of the slope of a Position vs. Time graph. Include a discussion of positive and negative slope.
- What type of motion is occurring when the slope of a Position vs. Time graph is zero?
- What type of motion is occurring when the slope of Position vs. Time graph is constant?
- What type of motion is occurring when the slope of a Position vs. Time graph is decreasing? What type of motion is occurring when the slope of a Position vs. Time graph is increasing?

AnalysisPART II – VELOCITY VS. TIME GRAPH

- What type of motion is occurring when the slope of a Velocity vs. Time graph is zero?
- What type of motion is occurring when the slope of a Velocity vs. Time graph is positive? What type of motion is occurring when the slope of a Velocity vs. Time graph is negative?
- What does the area under a Velocity vs. Time graph represent?

3rd pd. - 10/4 - Do now – on a new piece of paper

- The diagrams below represent two types motions. One is constant motion, the other, accelerated motion. Which one is constant motion and which one is accelerated motion? explain your answer.

10/5 do now Determine resultant graphically

- Vector A: 3 m East; Vector B: 6 m North
- You need to
- Indicate your scale
- show magnitude and direction of the resultant

objectives

- Vector addition – graphically and mathematically
- Homework
- Castle learning
- Vector HW 2

Steps for adding vectors using head and tail method

- Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
- Pick a starting location and draw the first vector to scale in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
- Starting from where the head of the first vector ends, draw the second vector to scale in the indicated direction. Label the magnitude and direction of this vector on the diagram.
- Repeat steps 2 and 3 for all vectors that are to be added
- Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as Resultant or simply R.
- Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
- Measure the direction of the resultant using the counterclockwise convention.

practice

- A man walks east for 3 meters, then south for 5 meters, then west for 6 meters.
- Draw his path in the area below using a scale of 1 centimeter = 1 meter.
- Draw the man’s final displacement vector.
- Measure the length of the vector on your paper.
- Calculate the man’s final displacement in meters
- Devise a way to solve this problem using your knowledge of geometry. Explain your method and show your work.
- How do the results of the two methods compare to one another?

The commutative property of vectors

A + B = B + A

example

- A 5.0-newton force and a 7.0-newton force act concurrently on a point. As the angle between the forces is increased from 0° to 180°, the magnitude of the resultant of the two forces changes from
- 0.0 N to 12.0 N
- 2.0 N to 12.0 N
- 12.0 N to 2.0 N
- 12.0 N to 0.0 N

example

- A 3-newton force and a 4-newton force are acting concurrently on a point. Which force could not produce equilibrium with these two forces?
- 1 N
- 7 N
- 9 N
- 4 N

example

- As the angle between two concurrent forces decreases, the magnitude of the force required to produce equilibrium
- decreases
- increases
- remains the same

10/9 do now

- Write all you know all about vector
- Definition:
- Examples (3):
- Representation:
- Ways to add vectors
- Head to tail: (sketch)
- Parallelogram method: (sketch)

objectives

- Homework questions?
- How to add vectors mathematically?
- Homework: castle learning
- No post session today
- Homework quiz is on Friday

Add vectors mathematicallyApply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector

The procedure is restricted to the addition of two vectors that make right angles to each other.

opp.

opp.

tanθ =

adj.

adj.

(

)

θ = tan-1

Using tangent function to determine a Vector's DirectionHyp.

opp.

θ

adj.

example

- Example: Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.

Note: The measure of an angle as determined through use of SOH CAH TOA is not always the direction of the vector.

R2 = (5.0km)2 + (10km)2

R = 11 km

Or at 26 degrees south of west

example

- An archaeologist climbs the Great Pyramid in Giza, Egypt. If the pyramid’s height is 136 m and its width is 2.30 x 102 m, what is the magnitude and the direction of the archaeologist’s displacement while climbing from the bottom of the pyramid to the top?

B

A

B

A

R

R

A

Equilibrant

Equilibrant

B

Equilibrant- The equilibrant vectors of A and B is the opposite of the resultant of vectors A and B.
- Example:

Head to tail

Parallelogram

Vector Components

- In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to transform the vector into two parts with each part being directed along the coordinate axes.

Any vector directed in two dimensions can be thought of as having an influence in two different directions.

- Each part of a two-dimensional vector is known as a component.
- The components of a vector depict the influence of that vector in a given direction.
- The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.
- The single two-dimensional vector could be replaced by the two components.

Vectors can be broken into COMPONENTS

X-Y system of components

AX = A cos θ

AY = A sin θ

Example

vi = 5.0 m/s at 30°

vix = 5.0 m/s (cos 30°) = 4.33 m/s

viy = 5.0 m/s (sin 30°) = 2.5 m/s

Any vector can be broken into unlimited sets of components

EXAMPLE

Calculate the x and y components of the following vectors.

a. A = 7 meters at 14°

b. B = 15 meters per second at 115°

c. C = 17.5 meters per second2 at 276°

Adding with Components

Vectors can be added together by adding their COMPONENTS

Results are used to find

RESULTANT MAGNITUDE

RESULTANT DIRECTION

Adding Vectors Algebraically

Example

Add vectors D and F by following the steps below.

a. Calculate the components of vectors D and F.

D = 35 meters at 25° F = 55 meters at 190°

b. Calculate the sum of the x-components of vectors D and F.

c. Calculate the sum of the y-components of vectors D and F.

d. Sketch the resultant x and y vectors on the axes below.

e. Calculate the length of the resultant generated by the resultant components

f. Calculate the direction of the resultant generated by the resultant components

Exit slip – hand in by the end of class

A bus heads 6.00 km east, then 3.5 km north, then 1.50 km at 45o south of west. What is the total displacement?

A: 6.0 km, 0° CCWB: 3.5 km, 90° CCWC: 1.5 km, 225° CCW

+

+

A

B

C

Cx = Ccos225o = -1.06 km

Cy = Csin225o = - 1.06 km

Relative Motion

and

2-D Kinematics

objectives

Homework – castle learning

Homework quiz – Friday

Lab – Thu. / Fri. – Dress warm

How would Homer know that he is hurtling

through interstellar space if his speed were

constant?

Without a window, he wouldn’t!

All of the Laws of Motion apply within his

FRAME of REFERENCE

ALL Motion is RELATIVE!

Do you feel like you are motionless

right now?

The only way to define motion is by changing position…

The question is changing position relative to WHAT?!?

MORE MOTION!!!

YOU'RE NOT!

You are moving at about 1000 miles per hour relative to the

center of the Earth!

The Earth is hurtling around the Sun at over

66,000 miles per hour!

Example #1

A train is moving east at 25 meters per second. A man on the train gets up and walks toward the front at 2 meters per second.

How fast is he going?

Depends on what we want to relate his speed to!!!

+2 m/s (relative to a fixed point on the train)

+27 m/s (relative to a fixed point on the Earth)

vperson = +2 m/s

vtrain = +25 m/s

Non-Parallel Vectors

What happens to the aircraft’s forward speed when the wind changes direction?

Wind is now slowing the plane

somewhat AND pushing it

SOUTH.

Wind is now working against

the aircraft thrust, slowing it

down, but causing no drift.

Wind is now NOT having any

effect on forward movement,

but pushes plane SOUTH.

Wind is still giving the plane

extra speed, but is also

pushing it SOUTH.

Wind in same direction

as plane – adds to overall

velocity!

No wind – plane

moves with velocity

that comes from engines

vthrust

vwind

example

- Which pair of forces acting concurrently on an object will produce the resultant of greatest magnitude?

100 km/hr)2 + (25 km/hr)2 = R2

10 000 km2/hr2 + 625 km2/hr2 = R2

10 625 km2/hr2 = R2

SQRT(10 625 km2/hr2) = R

103.1 km/hr = R

tan(θ) = (opposite/adjacent)

tan (θ) = (25/100)

θ = tan-1(25/100)

θ = 14.0o – CCW 256o

Analysis of a Riverboat's Motion

- The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat.

Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)?

(4.0 m/s)2 + (3.0 m/s)2 = R2

16 m2/s2 + 9 m2/s2 = R2

25 m2/s2 = R2

SQRT (25 m2/s2) = R

5.0 m/s = R

tan(θ) = (opposite/adjacent)

tan(θ) = (3/4)

θ = tan-1 (3/4)

θ = 36.9o

5 m/s

- If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?

time = distance /(ave. speed)

time = (80 m)/(4 m/s) = 20 s

80 m

5 m/s

- What distance downstream does the boat reach the opposite shore?

distance = ave. speed * time

d

distance = = (3 m/s) * (20 s)

distance = 60 m

Perpendicular Kinematics

Critical variable in multi dimensional problems is TIME.

We must consider each dimension SEPARATELY, using TIME as the only crossover VARIABLE.

Example

Now, assume that as the swimmer moves ACROSS the river, a current pushes him DOWNSTREAM at 0.1 meter per second.

200 m

vs = 0.5 m/s

How long will it take the swimmer

to get across?

t =0

vc= 0.1 m/s

The time to cross is unaffected! The

swimmer still arrives on the other bank

in 400 seconds. What IS different?

t = 400 s

The arrival POINT will be

shifted DOWNSTREAM!

A swimmer moving at 0.5 meters per second swims across a 200 meter wide river.

Exit slip

- Guided notes – pp 14 -15
- Please solve practice problems 1, 2, 3 on a separate sheet of paper and hand in by the end of the class.

Lab 7 - measuring immeaurable heights

- Purpose(5 points): To determine the heights of objects using trigonometry.
- Material(5 points): protractor, straw, tape measure.
- Objects to be measured: LCD projector (in class practice), Scoreboard, top of announcer box, footbal field ligth fixtrue, highschool roof

4 people groups

- 1 looker, 1 measurer (dist), 1 reader (angle), 1 recorders

10/11/12 do now

- A motorboat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.
- What is the resultant velocity of the motorboat?
- If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
- What distance downstream does the boat reach the opposite shore?

7 m/s

d = ?

4 m/s

objectives

- What is a Projectile?
- Characteristics of a Projectile's Trajectory

Homework – castle learning

question

- The long jumper builds up speed in the x-direction and jumps, so there is also a component of speed in the y direction. Does the angle of take-off matter to the jumper?

Ground Launched Projectiles

What is a projectile?

An object that is launched into the air with some INITIAL VELOCITY

Can be launched at ANY ANGLE

In FREEFALL after launch (no outside forces except force of gravity)

The path of the projectile is a PARABOLA

Initial velocity

What is the horizontal part of

the soccer ball’s initial velocity?

What is the vertical part of

the soccer ball’s initial velocity?

Pythagorean Theorem

vi2 = vix2 + viy2

vix = vi cos θ

viy = vi sin θ

12 m/s

6 m/s

30°

10.4 m/s

What do we know or assume about the vertical part of a projectile problem?

Initial vertical velocity = vi sin θ

Acceleration = -9.81 m/s2

Vertical speed will be 0 at the maximum height

Time to top = HALFtotal time in the air

Find time to top using final velocity equation

Vertical Distance – Max height

Use time to top and solve vertical distance equation

What do we know or assume about the horizontal part of a projectile problem?

Initial vertical velocity = vi cos θ

Acceleration = 0 (if we assume no air resistance)

Horizontal Distance – Range

Use total time and solve horizontal distance equation

Projectile Vector Diagram

t1/2

vfy = 0 (at top)

x y

a = 0 a = -9.81 m/s2

vix = vi cos θ viy = vi sin θ

vfy = 0 (at top)

ttot = 2t1/2

ttot

summary

- A projectile is any object upon which the only force is gravity
- Projectiles travel with a parabolic trajectory due to the influence of gravity,
- In horizontal direction, the projectile has no acceleration, its velocity is constant: vx = vicosθ
- In vertical direction, the projectile has acceleration: a = -9.81 m/s/s. Its velocity of a projectile changes by -9.81 m/s each second. Same as a free falling object.
- The horizontal motion of a projectile is independent of its vertical motion

Example

- An object was projected horizontally from a tall cliff. The diagram represents the path of the object, neglecting friction.
- Comparing the following at point A & B:
- Acceleration
- Horizontal velocity
- Vertical velocity

A projectile is fired with initial horizontal velocity at 10.00 m/s, and vertical velocity at 19.62 m/s. Determine the horizontal and vertical velocity at 1 – 5 seconds after the projectile is fired.

Lab 7 - measuring immeaurable heights

- Purpose(5 points): To determine the heights of objects using trigonometry.
- Material(5 points): protractor, straw, tape measure.
- Objects to be measured: LCD projector (in class practice), Scoreboard, top of announcer box, footbal field ligth fixtrue, highschool roof

4 people groups

- 1 looker, 1 measurer (dist), 1 reader (angle), 1 recorders

10/12 do now

- A motorboat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.
- What is the resultant velocity of the motorboat?
- If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
- What distance downstream does the boat reach the opposite shore?

7 m/s

d = ?

4 m/s

Exit slip – in complete sentences

- Describe following of a projectile
- The path
- Acceleration (magnitude and direction)
- Horizontal motion
- Graph horizontal velocity vs. time graph
- Vertical motion
- Graph vertical velocity vs. time
- The meaning of horizontal and vertical components are independent of each other
- The effect of air resistance on a projectile

objectives

- Homework quiz
- How to use equations to solve projectile problems?
- Homework: Regents Physics Work Sheets – Ch 3 – 2D motions and projectiles – pp 1-4
- Kinematics unit test is on next Friday – 10/19

Equation for projectile motion

d = ½ (vi + vf)t

vf = vi + at

d = vit + ½at2

vf2 = vi2 + 2ad

Vertical

- ay = -9.81 m/s2

Horizontal

- ax = 0

viy = visinθ

vfy = 0 (at top)

vfy = - viy (at same height)

y = ½ (viy + vfy)t

vfy = viy + ayt

y = viyt + ½ayt2

vfy2 = viy2 + 2ayy

vix = vicosθ

x = vix∙t

Initial Velocity Components

- Since velocity is a vector quantity, vector resolution is used to determine the components of velocity.

vi2 = vix2 + viy2

SOH CAH TOA

sinθ = viy / vi

viy = visinθ

cosθ = vix / vi

vix = vicosθ

vi

viy

θ

vix

Special case: horizontally launched projectile:

θ = 0o:viy = visinθ = 0; vix = vicosθ = vi

Examples

- Practice A: A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal.
- Practice B: A motorcycle stunt person traveling 70 mi/hr jumps off a ramp parallel to the horizontal.
- Practice C: A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal.

example

- A machine fired several projectiles at the same angle, θ, above the horizontal. Each projectile was fired with a different initial velocity, vi. The graph below represents the relationship between the magnitude of the initial vertical velocity, viy, and the magnitude of the corresponding initial velocity, vi, of these projectiles. Calculate the magnitude of the initial horizontal velocity of the projectile, vix, when the magnitude of its initial velocity, vi, was 40. meters per second.

The point of resolving an initial velocity vector into its two components is to use the values of these two components to analyze a projectile's motion and determine such parameters as

- the horizontal displacement,
- the vertical displacement,
- the final vertical velocity,
- the time to reach the peak of the trajectory,
- the time to fall to the ground, etc.

Example

- A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.

Due to the symmetrical nature of a projectile's trajectory: viy = - vfy

Solve for t – use vertical information:

vfy = viy + ayt

t = 3.61 s

Solve for x – use horizontal information:

x = vix•t + ½ •ax•t2

x = 63.8 m

Solve for peak height – use vertical information:

vfy2 = viy2 + 2ayypeak

At the top, vfy = 0

ypeak = 15.9 m

example

- A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

t = 1.1 s

x = 12.2 m

ypeak = 1.6 m

Example

- A cannon elevated at an angle of 35° to the horizontal fires a cannonball, which travels the path shown in the diagram. [Neglect air resistance and assume the ball lands at the same height above the ground from which it was launched.] If the ball lands 7.0 × 102 meters from the cannon 7.0 seconds after it was fired,
- what is the horizontal component of its initial velocity?
- what is the vertical component of its initial velocity?

Horizontal Projectiles

Horizontally Launched Projectiles

- A horizontally launched projectile is a projectile launched with only horizontal speed. Its initialvertical speed is zero.

Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.

As the red ball rolls off the edge, a green

ball is dropped from rest from the same

height at the same time

Which one will hit the ground first?

They will hit

at the SAME

TIME!!!

The same time?!? How?!?

The green ball falls from rest

and has no initial

velocity IN EITHER

DIRECTION!

viy and vix = 0

The red ball has an initial

HORIZONTAL velocity (vix)

But does not have any initial

VERTICAL velocity (viy = 0)

vix

One Dimension at a Time

Both balls begin with no VERTICAL VELOCITY

Both fall the same DISTANCE VERTICALLY

Find time of flight by solving in the appropriate dimension

We can find an object’s displacement in

EITHER DIMENSION using TIME

Example #1

A bullet is fired horizontally from a gun that is 1.7 meters above the ground with a velocity of 55 meters per second.

At the same time that the bullet is fired, the shooter drops an identical bullet from the same height.

Which bullet hits the ground first?

Both hit the ground at the same time

Velocity of Horizontal Projectiles

Horizontal motion is constant: velocity is constant.

Vertical: same as drop the ball from rest: velocity is increasing by 9.81 m/s every second

Displacement of horizontal Projectiles

Horizontal motion is constant: x = vix∙t

Vertical: same as free fall with initial zero velocity: y = ½ a∙t2 (a = - 9.81 m/s2)

Equation for horizontal projectile

Vertical

- ay = -9.81 m/s2

Horizontal

- ax = 0

θ = 0

d = ½ (vi + vf)t

vf = vi + at

d = vit + ½at2

vf2 = vi2 + 2ad

viy = = 0

y = ½ (viy + vfy)t

vfy = viy + ayt

y = viyt + ½ayt2

vfy2 = viy2 + 2ayy

vix = vicosθ = vi

x = vix∙t

x and y has the same t

Example #2

An airplane making a supply drop to troops behind enemy lines is flying with a speed of 300 meters per second at an altitude of 300 meters.

How far from the drop zone should the aircraft drop the supplies?

Need time from vertical

dy = viyt + ½ ayt2

300 m = 0 + ½ (-9.81 m/s2)t2

t = 7.82 s

Use time in horizontal

dx = vixt + ½ axt2

dx = (300 m/s)(7.82 s) + 0

dx = 2346 m

Example #3

A stuntman jumps off the edge of a 45 meter tall building to an air mattress that has been placed on the street below at 15 meters from the edge of the building.

What minimum initial velocity does he need in order to make it onto the air mattress?

Need time from vertical

dy = viyt + ½ ayt2

45 m = 0 + ½ (-9.81 m/s2)t2

t = 3.03 s

Use time to find v

v = d / t

v = 15 m / 3.03 s

v = 4.95 m /s

Example #4

A CSI detective investigating an accident scene finds a car that has flown off the edge of a cliff. The car is 79 meters from the edge of the 25 meter high cliff.

What was the car’s initial horizontal velocity as it went off the edge?

Example #4

Need time from vertical

dy = viyt + ½ ayt2

25 m = 0 + ½ (-9.81 m/s2)t2

t = 2.26 s

Use time in horizontal

dx = vixt + ½ axt2

79 m = vix (2.26 s) + 0

vix = 34.96 m/s

Example #5

The path of a stunt car driven horizontally off a cliff is represented in the diagram below. After leaving the cliff, the car falls freely to point A in 0.50 second and to point B in 1.00 second.

- Determine the magnitude of the horizontal component of the velocity of the car at point B. [Neglect friction.]

- Determine the magnitude of the vertical velocity of the car at point A.

- Calculate the magnitude of the vertical displacement, dy, of the car from point A to point B. [Neglect friction.]

Practice #1

- A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

Practice #2

- A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Lab Projectile motion

PURPOSE

- To predict the impact point of a ball in projectile motion.

MATERIALS

- Computer, books, Vernier computer interface, ruler, Logger Pro, two ring stands, two Vernier Photogates, two right-angle clamps, ball (1 to 5 cm diameter), meter stick, masking tape for target

PRELIMINARY QUESTIONS

- If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor? What assumptions must you make?
- If the ball in Question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.

3. A pair of computer-interfaced Photogates can be used to accurately measure the time interval for an object to break the beam of one Photogate and then another. If you wanted to know the velocity of the object, what additional information would you need?

Lab requirement

- Purpose (5)
- Material (5)
- Preliminary questions (30)
- Data section (35)
- Analysis questions (25)

Lab 7 - Vector Treasure Hunt

OBJECTIVES

- create a series of directions to lead to a specific object.
- follow directions to locate a specific object.
- develop standard notation for writing direction symbols.
- generate a scale map.

Materials

- Meter stick
- Index cards

10/10 do now

- Forces F1and F2act concurrently on point P, as shown in the diagram. What is the equilibrant of F1and F2? (include magnitude and direction)

10/11 do now

- A motorboat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.
- What is the resultant velocity of the motorboat?
- What distance downstream does the boat reach the opposite shore?

7 m/s

d = ?

4 m/s

10. N

10. N

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