Kinematics Power Point

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Kinematics in TwoDimensions

Kinematics in One Dimension (Review)

Table of Contents: Kinematics in 2DClick on the topic to go to that section

Adding Vectors in Two Dimensions

Vector ComponentsProjectile MotionGeneral Problems

Basic Vector Operations

Key Terms and EquationsKinematics Equations: 2-Dimensional Equations

v = v0 + a t vx = v cos(θ)v2 = v0

2 + 2 a Δx vy = v sin(θ)x = x0 + v0t + 1/2 a t2 v = √(vx2 + vy2)

θ = tan-1 (vy / vx)

Right Triangle Equations:

a2 + b2 = c2

SOH CAH TOAsin(θ) = Opposite/Hypotenusecos(θ) = Adjacent/Hypotenusetan(θ) = Opposite/Adjacent

Return toTable ofContents

Kinematics in One Dimension

Review of 1D Kinematics

• Kinematics is the description of how objects move with respect to a defined reference frame.

• Displacement is the change in position of an object.

• Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time.

• Instantaneous velocity is the limit as the time becomes infinitesimally short.

• Average acceleration is the change in velocity divided by the time.

Review of 1D Kinematics• Instantaneous acceleration is the limit as the time interval becomes infinitesimally small.

• There are four equations of motion for constant acceleration, each requires a different set of quantities.

v2 = vo2 + 2a(x xo)

x = xo + vot + ½at2v = vo + at

v = v + vo2

1 Starting from rest, you accelerate at 4.0 m/s2 for 6.0s. What is your final velocity?

v = vo + atv = 0 + 4(6)v = 24m/s

2 You have an initial velocity of 3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what is your final velocity?

v = vo + atv = 3 + 2.5(9)v = 19.5m/s

3 How much time does it take to come to rest if your initial velocity is 5.0 m/s and your acceleration is 2.0 m/s2?

v = vo + at0 = 5 + 2tt = 2.5s

4 An object moves at a constant speed of 6 m/s. This means that the object:

A Increases its speed by 6 m/s every second

B Decreases its speed by 6 m/s every second

C Doesn’t move

D Has a positive acceleration

E Moves 6 meters every second

5 A snapshot of three racing cars is shown on the diagram. All three cars start the race at the same time, at the same place and move along a straight track. As they approach the finish line, which car has the lowest average speed?

A Car IB Car II

C Car III

D All three cars have the same average speed

E More information is required

In physics there is another approach in addition to algebraic which is called graphical analysis.

The following formula v = v0 + at can be interpreted by the graph.

We just need to recall our memory from math classes where we already saw a similar formula y = mx + b.

Motion at Constant Acceleration

From these two formulas we can some analogies:

Velocity v ⇒ y (depended variable of x),v0 ⇒ b (intersection with vertical axis),t ⇒ x (independent variable), a ⇒ m ( slope of the graph the ratio between rise and run Δy/Δx).

y = m x + b

v = a t + v0 (or v = v0 + a t)



The formula a =Δv/Δt shows that the value of acceleration same as the slope on a graph of velocity versus time.

6 The velocity as a function of time is presented by the graph. What is the acceleration?

a = slope = Δv/Δt =(10 m/s 2 m/s)/40 s = 0.2 m/s2

7 The velocity as a function of time is presented by the graph. Find the acceleration.

a = slope = Δv/Δt = (0 m/s 25 m/s)/10 s = 2.5 m/s2

The acceleration graph as a function of time can be used to find the velocity of a moving object.

When the acceleration is constant it can be shown on the graph as a straight horizontal line.


In order to find the change in velocity for a certain limit of time we need to calculate the area under the acceleration versus time graph.


The change in velocity during first 12 seconds is equivalent to the shadowed area (4χ12 = 48).

The change in velocity during first 12 seconds is 48 m/s.

8 What is the velocity of the object at 5 s?

A 1 m/sB 2 m/s

C 3 m/s

D 4 m/sE 5 m/s

9 Which of the following statements is true?

A The object speeds up

B The object slows down

C The object moves with a constant velocity

D The object stays at rest

E The object is in free fall

10 The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds?

Δv = area = (3χ10) = 30 m/s

11 The graph represents the relationship between velocity and time for an object moving in a straight line. What is the traveled distance of the object at 9 s?

A 10 m

B 24 m

C 36 m

D 48 m

E 56 m

12 Which of the following is true?

A The object increases its velocity

B The object decreases its velocity

C The object’s velocity stays unchanged

D The object stays at rest


13 What is the initial position of the object?

A 2 m

B 4 m

C 6 m

D 8 m

E 10 m

14 What is the velocity of the object?

A 2 m/s

B 4 m/s

C 6 m/s

D 8 m/sE 10 m/s

15 What is the initial position of the object?

A 2 m

B 4 m

C 6 m

D 8 m

E 10 m

16 The graph represents the position as a function of time of a moving object. What is the velocity of the object?

A 5 m/s

B 5 m/s

C 10 m/s

D 10 m/s

E 0 m/s

Free FallAll unsupported objects fall towards the earth with the same acceleration.

We call this acceleration the "acceleration due to gravity" and it is denoted by g.

g = 9.8 m/s2

Keep in mind, ALL objects accelerate towards the earth at the same rate.

g is a constant!

It speeds up(negative acceleration)g = 9.8 m/s2

It stops momentarily.v = 0g = 9.8 m/s2

An object is thrown upward with initial velocity, vo

It slows down.(negative acceleration)g = 9.8 m/s2What happens when it goes up?

What happens when it goes down?

What happens at the top?

It returns with itsoriginal velocity.What happens when it lands?

It speeds up.(negative acceleration)g = 9.8 m/s2



It slows down.(negative acceleration)

g = 9.8 m/s2

It returns with itsoriginal velocity.

a

v0

On the way up:

a

v1

v1a

v2

v2

a

a

va

av0

On the way down:

v1

v1v2

v2

v

vt = 0 s

t = 1 s

t = 2 s

t = 3 s t = 0 s

t = 1 s

t = 2 s

t = 3 s

v(m/s)

t (s)



It returns with itsoriginal velocity but in the opposite direction.

For any object thrown straight up into the air, what does the velocity vs time graph look like?

E

17 An object moves with a constant acceleration of 5 m/s2. Which of the following statements is true?

A The object’s velocity stays the same

B The object moves 5 m each second

C The object’s acceleration increases by 5 m/s2 each second

D The object’s acceleration decreases by 5 m/s2 each second

E the object’s velocity increases by 5 m/s each second

18 A truck travels east with an increasing velocity. Which of the following is the correct direction of the car’s acceleration?

A

B

C

D

E

A

19 A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the final speed of the car compared to the truck?

A Half as much

B The same

C Twice as much

D Four times as much

E One quarter as much

C

20 A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the traveled distance of the car compared to the truck?

A Half as much

B The same

C Twice as much



D

21 A modern car can develop an acceleration four times greater than an antique car like “Lanchester 1800”. If they accelerate over the same distance, what would be the velocity of the modern car compared to the antique car?

A Half as much

B The same

C Twice as much



C

22 An object is released from rest and falls in the absence of air resistance. Which of the following is true about its motion?

A Its acceleration is zero

B Its acceleration is constant

C Its velocity is constant

D Its acceleration is increasing

E Its velocity is decreasing

B

23 A ball is thrown straight up from point A it reaches a maximum height at point B and falls back to point C. Which of the following is true about the direction of the ball’s velocity and acceleration between A and B?

A

B

C

D

E

B

24 A ball is thrown straight up from point A it reaches a maximum height at point B and falls back to point C. Which of the following is true about the direction the ball’s velocity and acceleration between B and C?

A

B

C

D

E

D

25 A football, a hockey puck, and a tennis ball all fall down in the absence of air resistance. Which of the following is true about their acceleration?

A The acceleration of the football is greater than the other two

B The acceleration of the hockey puck is greater than the other two

C The acceleration of the tennis ball is greater than the other two

D They all fall down with the same constant acceleration


D

26 A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is H and in the second trial 4H. Compare the time it takes for the package to reach the surface in the second trial to that in the first trial?

A The time in the second trial is four times greater

B The time in the second trial is two times greater

C The time the same in both trials because it doesn’t depend on height

D The time in the second trial is four times less

E The time in the second trial is two times less

B

27 Two baseballs are thrown from the roof of a house with the same initial speed, one is thrown up, and the other is down. Compare the speeds of the baseballs just before they hit the ground.

A The one thrown up moves faster because the initial velocity is up

B The one thrown down moves faster because the initial velocity is down

C They both move with the same speed

D The one thrown up moves faster because it has greater acceleration

E The one thrown down moves faster because it has greater acceleration

B

28 An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is v0 and second time he increases the initial sped to 4v0. How would you compare the maximum height in the first trial to that in the first trial?

A Two times greater

B Four times greater

C Eight times greater

D Sixteen times greaterE The same

D

29 An acorn falls from an oak tree. You note that it takes 2.5 seconds to hit the ground. How fast was it going when it hit the ground?

v = vo + atv=0(9.8)2.5v=24.5 m/s

30 An arrow is fired into the air and it reaches its highest point 3 seconds later. What was its velocity when it was fired?

v = vo + at0 = vo + 9.8(3)vo = 29.4m/s

31 You accelerate from 20m/s to 60m/s while traveling a distance of 200m; what was your acceleration?

v2 = vo2 + 2aΔx602 = 202 + 2a(200)a = 8m/s2

32 A dropped ball falls 8.0m; what is its final velocity?

v2 = vo2 + 2aΔyv2 = 2aΔyv2 = 2(9.8)8v = 12.5m/s

33 A ball with an initial velocity of 25m/s is subject to an acceleration of 9.8 m/s2; how high does it go before coming to a momentary stop?

v2 = vo2 + 2aΔy0 = 252 + 2(9.8)Δyy = 31.9m

34 A water drop falls down from the roof of a house during 3 s, how high is the house?

x = 1/2 (9.8) (3)2\x = 44 m

35 A ball is thrown vertically down from the edge of a cliff with a speed of 8 m/s, how high is the cliff, if it took 6 s for the ball to reach the ground?

x = (8)(60 + 1/2(9.8)(6)2x = 224.4 m

36 What is the landing velocity of an object that is dropped from a height of 49 m?

x = 31 m/s

37 An object is thrown vertically up with a velocity of 35 m/s. What was the maximum height it reached?

x = 62.5 m

38 A boy throws a ball vertically up and catches it after 3 s. What height did the ball reach?

x = 11 m

Vectors vs. ScalarsWe used vectors last year.You'll recall that:

• Vectors have a magnitude AND a direction

• Scalars only have a magnitude

39 Which one of the following is an example of a vector quantity?

A distance B velocity C mass D area

40 Which of the following is a vector quantity?

A SpeedB Time

C Traveled distance

D VelocityE Area

41 A runner runs halfway around a circular path of radius 10 m. What is the displacement of the jogger?

A 0 mB 10 m

C 20 mD 31.4 m

42 A runner runs halfway around a circular path of radius 10 m. What is the total traveled distance of the jogger?

A 0 mB 10 m

C 20 mD 31.4 m


Adding Vectors in Two Dimensions

Adding VectorsLast year, we learned how to add vectors along a single axis. The example we used was for adding two displacements.

West

North

South

East


1. Draw the first vector, beginning at the origin, with its tail at the origin.

West

North

South

East

y



2. Draw the second vector with its tail at the tip of the first vector.

North

West

South

East

West

North

South

East




3. Draw the Resultant (the answer) from the tail of the first vector to the tip of the last.

x

+y

y

+x

Adding VectorsThe direction of each vector matters.

In this first case, the vector sum of:

5 units to the East plus 2 units to West is

3 units to the East.

x

+y

y

+x

Adding VectorsThe direction of each vector matters. For instance, if the second vector had been 2 units to the EAST (not west), we get a different answer.

In this second case, the vector sum of:

5 units to the East plus 2 units to EAST is

7 units to the East.

Adding Vectors in 2DBut how about if the vectors are along different axes.

For instance, let's add vectors of the same magnitude, but along different axes.

What is the vector sum of:

5 units East plus

2 units North

West

North

South

East

Adding Vectors1. Draw the first vector, beginning at the origin, with its tail at the origin.

West

North

South

East



West

North

South

East



3. Draw the Resultant (the answer) from the tail of the first vector to the tip of the last.

West

North

South

East

Adding VectorsDrawing the Resultant is the same as we did last year.

But calculating its magnitude and direction require the use of right triangle mathematics. West

North

South

East

We know the length of both SIDES of the triangle (a and b), but we need to know the length of the HYPOTENUSE (c).

a

bc

Magnitude of A ResultantThe magnitude of the resultant is equal to the length of the vector.

We get the magnitude of the resultant from the Pythagorean Theorem:

c2 = a2 + b2

or in this case:

R2 = 52 + 22 R2 = 25 + 4 R2 = 29

R = √(29) = 5.4 units

West

North

South

Easta = 5 units

b = 2 unitsc = ?

43 Starting from the origin, a person walks 6 km east during first day, and 3 km east the next day. What is the net displacement of the person from the initial point in two days?

A 6 km, westB 3 km, east

C 10 km, east

D 5 km, westE 9 km, east

44 Starting from the origin, a car travels 4 km east and then 7 km west. What is the net displacement of the car from the initial point?

A 3 km, westB 3 km, east

C 4 km, east

D 7 km, westE 7 km east

45 Starting from the origin, a person walks 8 km east during first day, and 5 km west the next day. What is the net displacement of the person from the initial point in two days?

A 6 km, east

B 3 km, east

C 10 km, west

D 5 km, westE 9 km, east

46 What is the magnitude of the Resultant of two vectors A and B, if A = 8.0 units north and B = 4.5 units east?

9.18 units

47 What is the magnitude of the Resultant of two vectors A and B, if A = 24.0 units east and B = 15.0 units south?

28.3 units

Adding VectorsIn physics, we say the direction of a vector is equal to the angle θ between a chosen axis and the resultant.

In Kinematics, we will primarily use the xaxis to measure θ.

However, if we were to change the axis we used and apply the proper mathematical techniques, we should get the same result!

West

North

South

Eastθ

Adding VectorsTo find the value of the angle θ, we need to use what we already know:

the length of the two sides opposite and adjacent to the angle.

(remember SOH CAH TOA)

tan(θ) = opposite adjacent

West

North

South

Eastθ

Adding Vectors

tan(θ) = opp / adj

tan(θ) = (2 units) / (5 units)

tan(θ) = 2/5

tan(θ) = 0.40

To find the value of θ, we must take the inverse tangent:

θ = tan1(0.40) = 220

West

North

South

Eastθ

Adding Vectors

West

North

South

Eastθ

The Resultant is 5.4 units in the direction of 220 North of East

magnitude direction

48 What is the direction of the Resultant of the two vectors A and B if:

A = 8.0 units north and B = 4.5 units east?

θ = 60o

49 What is the direction of the Resultant of the two vectors A and B if:

A = 24.0 units east and B = 15.0 units south?

θ = 32o

50 Find the magnitude and direction of the resultant of two vectors A and B if:

A = 400 units north B = 250 units east

Magnitude = ?

471 units

51 Find the magnitude and direction of the resultant of two vectors A and B if:

A = 400 units north B = 250 units east

Direction = ?

θ = 58o

52 A student walks a distance of 300 m East, then walks 400 m North. What is the magnitude of the net displacement?

A 300 m

B 400 m

C 500 m

D 700 m

53 A student walks a distance of 300 m East, then walks 400 m North. What is the total traveled distance?

A 300 m

B 400 m

C 500 m

D 700 m

54 Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude of the sum

A is 2.0 m

B could be as small as 2.0 m, or as large as 12 mC is 12 m

D is larger than 12 m

55 The resultant of two vectors is the largest when the angle between them is

A 0°

B 45°C 90°

D 180°

56 The resultant of two vectors is the smallest when the angle between them is

A 0°B 45°C 90°

D 180°


Basic Vector Operations

West

North

South

Adding VectorsAdding Vectors in the opposite order gives the same resultant.

V1 + V2 = V2 + V1

West

North

South

East EastV1

V1

V2 V2

Adding VectorsEven if the vectors are not at right angles, they can be added graphically by using the "tail to tip" method.

The resultant is drawn from the tail of the first vector to the tip of the last vector.

V1 V2V3++ =

V1

V2

V3

VR

Adding Vectors...and the order does not matter.

V1

V3 V2 V3

V1

V2

+ + =VR

V3V1

V2

VR

Try it.

Adding VectorsVectors can also be added using the parallelogram method.

V1 + =V2

V1

V2

VR

Subtracting VectorsIn order to subtract a vector, we add the negative of that vector. The negative of a vector is defined as that vector in the opposite direction.

V1 =V2

VR

V1 + V2

V1

V2

=

Multiplication of Vectors by ScalarsA vector V can be multiplied by a scalar c. The result is a vector cV which has the same direction as V. However, if c is negative, it changes the direction of the vector.

V 2V ½V

57 Which of the following operations will not change a vector?

A Translate it parallel to itselfB Rotate itC Multiply it by a constant factorD Add a constant vector to it


Vector Components

Adding Vectors by ComponentsAny vector can be described as the sum as two other vectors called components. These components are chosen perpendicular to each other and can be found using trigonometric functions.

x

y

V

Vx

Vy

θ

Adding Vectors by ComponentsIn order to remember the right triangle properties and to better identify the functions, it is often convenient to show these components in different arrangements (notice vy below).

x

y

V

Vx

Vy

θ

58 Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis. The y component of A is

A 6.9 mB 6.9 mC 4.0 mD 4.0 m

59 Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis. The X component of A is

A 6.9 mB 6.9 mC 4.0 mD 4.0 m

60 If a ball is thrown with a velocity of 25 m/s at an angle of 37° above the horizontal, what is the vertical component of the velocity?

A 12 m/s B 15 m/s C 20 m/s D 25 m/s

61 If a ball is thrown with a velocity of 25 m/s at an angle of 37° above the horizontal, what is the horizontal component of the velocity?

A 12 m/s B 15 m/s C 20 m/s D 25 m/s

62 If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north of east have you walked?

A 19°

B 45°C 60°D 71°

63 A butterfly moves with a speed of 12.0 m/s. The x component of its velocity is 8.00 m/s. The angle between the direction of its motion and the x axis must be

A 30.0°. B 41.8°. C 48.2°. D 53.0°.

64 A 400m tall tower casts a 600m long shadow over a level ground. At what angle is the Sun elevated above the horizon?

A 34° B 42° C 48° D can't be found; not enough information

Adding Vectors by ComponentsUsing the tiptotail method, we can sketch the resultant of any two vectors.

x

y

V1

V2V

But we cannot find the magnitude of 'v', the resultant, since v1 and v2 are twodimensional vectors

Adding Vectors by ComponentsWe now know how to break v1 and v2 into components...

x

y

V1

V1x

V1y

V2

V2y

V2x

Adding Vectors by ComponentsAnd since the x and y components are one dimensional, they can be added as such.

x

y

V1x

V1y

V2y

V2x

Vx = v1x + v2x

Vy = v1y + v2y

65 Two vectors A and B have components (0, 1) and (1, 3), respectively. What are the components of the sum of these two vectors?

A (1, 4) B (1, 4) C (1, 2) D (1, 2)

66 Two vectors A and B have components (0, 1) and (1, 3), respectively. What is magnitude of the sum of these two vectors?

A 2.8 B 3.2 C 3.9 D 4.1

67 Vector A = (1, 3). Vector B = (3, 0). Vector C = A + B . What is the magnitude of C? A 3B 4C 5D 7

Adding Vectors by Components

V1

V2

1. Draw a diagram and add the vectors graphically.


x

y

V1

V2


2. Choose x and y axes.


x

y

V1

V1x

V1y

V2

V2y

V2x



3. Resolve each vector into x and y components.


x

y

V1

V1x

V1y

V2

V2y

V2x




4. Calculate each component.

v1x = v1 cos(θ1) v2x = v2 cos(θ2)

v1y = v1 sin(θ1) v2y = v2 sin(θ2)


x

y

V1

V1x

V1y

V2

V2y

V2x

Vx

Vy





5. Add the components in each direction.


x

y

V1

V1x

V1y

V2

V2y

V2x

V

Vx

Vy





5. Add the components in each direction.

6. Find the length and direction of the resultant vector.

v = √(vx2 + vy2) (length)

θ = tan1 ( vy ) (direction) vx

Example:

Graphically determine the resultant of the following three vector displacements:

1. 24m, 30º north of east

2. 28m, 37º east of north

3. 20m, 50º west of south

y N

x

E

S

W 30º

d1 = 24m

d1x

d1y

1. 24m, 30º north of east

Find the x and y components of the vector d1

y N

x

E

S

W

37ºd2 = 28m

d2x

d2y

2. 28m, 37º east of north


y N

x

E

S

W

50ºd3 =20m

d3x

d3y

3. 20m, 50º west of south


x (m) y (m)d1 20.8 12.0 d2 16.9 22.4 d3 15.3 12.9 Ʃ 22.4 21.5 tan θ = dy / dx

tan θ = 21.5 / 22.4

tan θ = 0.96

θ = tan1 (0.96)

θ = 43.8º

Using the results from each set of vector components, we can create a table to find the resultant vector components and the magnitude and direction:

d = √(dx2 + dy2)

d = √(22.42 + 21.52)d = 31 m

Graphically determine the magnitude and direction of the resultant of the following three vector displacements:

d1. 15 m, 30º north of east

d2. 20 m, 37º north of east

d3. 25 m, 45o north of east

xcomponent ycomponent

d1

d2

d3

Σ

68 Graphically determine the magnitude and direction of the resultant of the following three vector displacements:

1. 15 m, 30º north of east


3. 25 m, 45o north of east

Magnitude = ?

59.7 m

69 Graphically determine the magnitude and direction of the resultant of the following three vector displacements:



3. 25 m, 45o north of east

Direction = ?

38.9o

70 Which of the following is an accurate statement?

A A vector cannot have a magnitude of zero if one of its components is not zero.

B The magnitude of a vector can be equal to less than the magnitude of one of its components.

CIf the magnitude of vector A is less than the magnitude of vector B, then the xcomponent of A must be less than the xcomponent of B.

D The magnitude of a vector can be either positive or negative.


Projectile Motion

Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity. Its path is a parabola.

Vertical fall

Projectile Motion

vx

vy v

a = g

Projectile MotionProjectile motion can be understood by analyzing the vertical and horizontal motion separately.

The speed in the xdirection is constant.

The speed in the ydirection is changing.

Vertical fall

Projectile Motion

vx

vy v

a = g

A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land?

7.5 m

4.5 m/s

x = ?

x

x0 = 0vx0 = 4.5 m/sax = 0x = ?

y

y0 = 7.5 mvy0 = 0ay = 9.8 m/s2y = 0


7.5 m

4.5 m/s

x = ?

x

x0 = 0vx0 = 4.5 m/sax = 0x = ?

y

y0 = 7.5 mvy0 = 0ay = 9.8 m/s2y = 0

y = y0 + vy0t + ½ ay t2

0 = y0 + ½ ay t2

t = √(2y0/ay)

t = √(2*7.5m/s / 9.8m/s2)

t = 1.24 s


7.5 m

4.5 m/s

x = ?

x

x0 = 0vx0 = 4.5 m/sax = 0x = ?

t = 1.24 s

x = x0 +vx0t + ½at2

x = vx0t

x = (4.5m/s)(1.24s)

x = 5.58 m

y

y0 = 7.5 mvy0 = 0ay = 9.8 m/s2y = 0

y = y0 + vy0t + ½ at2

0 = y0 + ½ at2

t = √(2y0/ay)

t = √(2*7.5m/s / 9.8m/s2)

t = 1.24 s

A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land.

http://phet.colorado.edu/simulations/sims.php?sim=Projectile_Motion

http://phet.colorado.edu/simulations/sims.php?sim=projectile_motion

x

x0 = 0vx0 = 20 m/sa = 0x = ?

y

y0 = 15 mvy0 = 0a = g = 9.8 m/s2y = 0




x

x0 = 0vx0 = 20 m/sa = 0x = ?

t = 1.75 s

x = x0 +vx0t + ½at2

x = vx0t

x = (20m/s)(1.75s)

x = 34 m

y

y0 = 15 mvy0 = 0a = g = 9.8 m/s2

y = 0

y = y0 + vy0t + ½ at2

y0 + ½ at2

t = √(2y0/a)

t = √(2*15m/9.8m/s2)

t = 1.75 s




vx

vyv

vx

vy v

vx = v

vx

vyv

vy

vx

v

Projectile MotionIf an object is launched at an angle with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

Projectile Motion Solving Problems

1. Read the problem carefully.

2. Draw a diagram.

3. Choose an origin and coordinate system.

4. Choose a time interval which is the same for both directions.

5. List known and unknown quantities in both directions. Remember vx never changes and vy at the top is zero.

6. Plan how you will proceed. Choose your equations.

vx

vyv

vx

vy v

vx = v

vx

vyv

vy

vx

v

Projectile motion can be described by two kinematics equations:

horizontal component: vertical component:

x = x0 + v0xt + ½axt2 y = y0 + v0yt + ½ayt2

vx = v0x + axt vy = v0y + ayt

vx

vyv

vx

vy v

vx = v

vx

vyv

vy

vx

v

We can simplify these equations for each problem.

For example:

x direction y direction

x0 = 0 y0 = 0v0x = v0cosθ v0y=v0sinθax =0 ay = gx = v0cosθt vx = v0cosθ

y = v0sinθt 1/2gt2vy = v0sinθ gt

Flying time:

y = v0sinθt ½gt2 and y = 0

0 =v0sinθt ½gt2

Now we solve for time:

t = (2v0sinθ)/g

vx

vyv

vx

vy v

vx = v

vx

vyv

vy

vx

v

vx

vyv

vx

vy v

vx = v

vx

vyv

vy

vx

v

Horizontal range:

In order to find the horizontal range we need to substitute flying time in x(t).

x = v0cosθt

x = v0cosθ2v0sinθ/g

x = 2 v02cosθsinθ/g or x = v02sin2θ/g

vx

vyv

vx

vy v

vx = v

vx

vyv

vy

vx

v

Maximum height:

In order to find the maximum height we need to substitute a half of flying time in y(t).

y = v0sinθt ½gt2

y = v0sinθ(2v0sinθ/g/2) ½g(2v0sinθ/g/2)2

y = v02sin2θ/g v02sin2θ/2g

y = v02sin2θ/2g

71 Ignoring air resistance, the horizontal component of a projectile's velocity

A is zeroB remains constantC continuously increasesD continuously decreases

72 A ball is thrown with a velocity of 20 m/s at an angle of 60° above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory?

A 10 m/s B 17 m/s C 20 m/s D zero

73 Ignoring air resistance, the horizontal component of a projectile's acceleration

A is zeroB remains a nonzero constantC continuously increasesD continuously decreases

74 At what angle should a watergun be aimed in order for the water to land with the greatest horizontal range?

A 0°B 30°

C 45°D 60°

75 An Olympic athlete throws a javelin at four different angles above the horizontal, each with the same speed: 30°, 40°, 60°, and 80°. Which two throws cause the javelin to land the same distance away?

A 30° and 80° B 30° and 70° C 40° and 80° D 30° and 60°

76 You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be

A 1.4 times as muchB half as muchC twice as muchD four times as much

77 A ball is thrown at an original speed of 8.0 m/s at an angle of 35° above the horizontal. What is the speed of the ball when it returns to the same horizontal level?

A 4.0 m/s B 8.0 m/s C 16.0 m/s D 9.8 m/s

78 When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed?

A It is zeroB It is less than its initial speedC It is equal to its initial speedD It is greater than its initial speed

79 A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below?

A It is impossible to tell from the information givenB the stone C the ball D Neither, since both are traveling at the same speed

80 A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?

A 100 mB 170 mC 180 mD 210 m

A projectile is fired with an initial speed of 30 m/s at an angle of 30o above the horizontal.

a. Determine the total time in the air.b. Determine the maximum height reached by the projectile.c. Determine the maximum horizontal distance covered by the

projectile.d. Determine the velocity of the projectile 2 s after firing.


a. Determine the total time in the air.

vy-top = vy0 + ay ttop

ttop = (vtop - vy0) ay

ttop = (0m/s - 15m/s) 9.8m/s2

ttop = 1.5 s ttotal = 2 ttop

ttotal = 3 s

v0 = 30 m/sθ = 300

x yvx = v cosθ v0y = v sinθ

vy-top = 0 m/say = -9.8 m/s2

t = ?

v0y = v sin (θ)v0y = 30 sin(30)v0y = 15 m/s


b. Determine the maximum height reached by the projectile.

v0 = 30 m/sθ = 300

x yvx = v cosθax = 0

v0y = 15 m/svy-top = 0 m/say = -9.8 m/s2

ttotal = 3 s

ttop = 1.5s (from previous)

y = y0 + v0yt + 1/2 ay t2

y = 0 + (15)(1.5) + 1/2(-9.8)(1.5)2

y = 22.5m - 11.m

y =11.5 m


c. Determine the maximum horizontal distance covered by the projectile.

v0 = 30 m/sθ = 300



y = 11.5 m

ttotal = 3 s

vx = v cos θ

vx = 30 cos (300)

vx = 26 m/s

x = x0 + v0xt + 1/2 ax t2

x = 0 + v0xt + 0

x = v0xt

x = (26 m/s)(3 s)

x = 78 m


d. Determine the velocity of the projectile 2 s after firing.

x yvx = 26 m/sax = 0


y = 11.5 m

v0 = 30 m/sθ =300

ttotal = 3 s vf = √(vxf2 + vyf2)

vxf = vx0 = 26 m/s

vyf = vy0 + a t

vyf = 15 + (-9.8 m/s2)(2s)

vyf = -4.6 m/s

vf = √[ (26)2 +(-4.6)2 ]

vf = √(676 + 21.16)

vf = 26.4 m/s

magnitude direction

tan (θ) = (vy/vx)

tan (θ) = 4.6/26

θ = tan-1(4.6/26)

θ = 100 below the horizontal

12. A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45o above the horizontal.

a. Determine the maximum height reached by the projectile.b. Determine the total time in the air.c. Determine the maximum horizontal distance covered by the

projectile.d. Determine the velocity of the projectile just before it hits the

bottom of the cliff.


Δx

Δy

y0 = 200 m

v0 =30 m/s

θ = 450


v0y = v sin θvy-top = 0 m/say = -9.8 m/s2

y0 = 200 myf = 0m


a. Determine the maximum height reached by the projectile.



y0 = 200 myf = 0m

vy2 = v0y

2 + 2 a (y - y0)

vy2 - v0y

2 = 2a (y - y0)

vy2 - v0y

2 = (y - y0) 2a

y = (vy2 - v0y

2) + y0

2a

y = [0 - (30 sin(450)2]/-9.8 + 200y = 46m + 200my = 246m


b. Determine the total time in the air.



y0 = 200 myf = 0mytop = 246 m

tair = ttop + tbottom

ttop:

vytop = v0y + a ttop

ttop = (vy-top - v0y)/a

ttop = (0 - v sinθ) / a

ttop = [- 30 sin(45)] / - 9.8

ttop = 2.16s

tbottom:

ybot = ytop + vy-topt + 1/2 ay tbot2

ybot = ytop + 1/2 ay tbot

2

tbot = √[ (ybot - ytop) ] 2a

tbottom = √[(0 - 246)/(2*-9.8)]

tbottom = 3.54 s

tair = 2.16 s + 3.54stair = 5.7 s





y0 = 200 myf = 0mytop = 246 m

x = x0 + v0xt + 1/2 ax t2

x = v0xt

x = v cosθ t

x = 30 cos(45) (5.7s)

x = 121 m

tair = 5.7 s


d. Determine the velocity of the projectile just before it hits the bottom of the cliff.

vf = √(vxf2 + vyf2)

vxf = vx0 = 21.2 m/s

vyf = vy0 + a t

vyf = 21.2 + (-9.8 m/s2)(5.7)

vyf = -34.7 m/s

vf = √[ (21.2)2 +(-34.7)2 ]

vf = √(449 + 1205)

vf = 40.7 m/s

magnitude directiontan (θ) = (vy/vx)

tan (θ) = -34.7/21.2

tan(θ) = 1.64

θ = tan-1(1.64)

θ = -58.60



y0 = 200 myf = 0mytop = 246 m

tair = 5.7 s


General Problems

81 A bicyclist moves a long a straight line with an initial velocity v0 and slows down. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration?

Initial position Initial velocity Acceleration A Positive Negative Negative

B Positive Positive Negative

C Negative Positive NegativeD Negative Negative Positive

E Negative Negative Negative

82 A projectile is fired at 70 above the horizontal line with an initial velocity v0. At which of the following angles the projectile will land at the same distance as it is landed in the first trial?

A 20 B 30 C 40 D 45 E 50

1. A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s2.

a. How far will the motorcycle travel before catching the car?

b. How fast will it be going at that time?

c. How does that compare to the car’s velocity?

d. Draw the following graphs for the car: x(t), v(t), a(t).

e. Draw the following graphs for the motorcycle: x(t), v(t), a(t).

f. Write the equation of motion for the car.

g. Write the equation of motion for the motorcycle.


a. How far will the motorcycle travel before catching the car?


b. How fast will it be going at that time?.


c. How does that compare to the car’s velocity?


d. Draw the following graphs for the car: x(t), v(t), a(t).


e. Draw the following graphs for the motorcycle: x(t), v(t), a(t).


f. Write the equation of motion for the car.


g. Write the equation of motion for the motorcycle.

2. A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart.

a. Indicate every time interval for which speed (magnitude of the velocity) of the cart is decreasing.

b. Indicate every time at which the cart is at rest.

c. Determine the horizontal position x of the cart at t = 4 s if the cart is located at x0 = when t0 = 0.

d. Determine the traveled distance of the cart over 10 s from the beginning.

e. Determine the average speed of the cart for this time interval.

f. Find the acceleration of the cart during time: 0 s 4 s, 4 s 8 s, 8 s 10 s, 10 s 14 s, 14 s 16 s, 16 s 20 s.

g. On the axes below, sketch the acceleration graph for the motion of the cart from t = 0 s to t = 20 s.


a. Indicate every time interval for which speed (magnitude of the velocity) of the cart is decreasing.t = 0 s to t = 20 s.


b. Indicate every time at which the cart is at rest.


c. Determine the horizontal position x of the cart at t = 4 s if the cart is located at x0 = when t0 = 0.


d. Determine the traveled distance of the cart over 10 s from the beginning.


e. Determine the average speed of the cart for this time interval.


f. Find the acceleration of the cart during time: 0 s 4 s, 4 s 8 s, 8 s 10 s, 10 s 14 s, 14 s 16 s, 16 s 20 s.


g. On the axes below, sketch the acceleration graph for the motion of the cart from t = 0 s to t = 20 s.

3. Find the magnitude and the direction of vector C for the following cases.

a. A = 10 N at 0o, B = 20 N at 0o, C = A + Bb. A = 10 N at 0o, B = 20 N at180o, C = A + Bc. A = 10 N at 180o, B = 20 N at 180o, C = A + Bd. A = 10 N at 0o, B = 20 N at 90o, C = A + Be. A = 10 N at 90o, B = 20 N at 0o, C = A + B


a. A = 10 N at 0o, B = 20 N at 0o, C = A + B


b. A = 10 N at 0o, B = 20 N at180o, C = A + B


c. A = 10 N at 180o, B = 20 N at 180o, C = A + B


d. A = 10 N at 0o, B = 20 N at 90o, C = A + B


e. A = 10 N at 90o, B = 20 N at 0o, C = A + B

4. Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps.

a. D = 10 N at 37o Find Dx and Dy.b. E = 20 N at 25o. Find Ex and Ey.c. Find Gx = Dx + Exd. Find Gy = Dy + Eye. Find the magnitude of G from its componentsf. Find the direction of G.


a. D = 10 N at 37o Find Dx and Dy.


b. E = 20 N at 25o. Find Ex and Ey.


c. Find Gx = Dx + Ex


d. Find Gy = Dy + Ey


e. Find the magnitude of G from its components


f. Find the direction of G.


a. A = 40 N at 0o, B = 10 N at 0o, C = A + Bb. A = 40 N at 0o, B = 10 N at 180o, C = A + Bc. A = 40 N at 180o, B = 10 N at 180o, C = A + Bd. A = 40 N at 0o, B = 10 N at 90o, C = A + Be. A = 40 N at 90o, B = 10 N at 0o, C = A + B


a. A = 40 N at 0o, B = 10 N at 0o, C = A + B


b. A = 40 N at 0o, B = 10 N at 180o, C = A + B


c. A = 40 N at 180o, B = 10 N at 180o, C = A + B


d. A = 40 N at 0o, B = 10 N at 90o, C = A + B


e. A = 40 N at 90o, B = 10 N at 0o, C = A + B


a. D = 30 N at 65o. Find Dx and Dy.b. E = 45 N at 15o. Find Ex and Ey.c. Find Gx = Dx + Exd. Find Gy = Dy + Eye. Find the magnitude of G from its componentsf. Find the direction of G.


a. D = 30 N at 65o. Find Dx and Dy.


b. E = 45 N at 15o. Find Ex and Ey.


c. Find Gx = Dx + Ex


d. Find Gy = Dy + Ey


e. Find the magnitude of G from its components


f. Find the direction of G.

7. Two forces 300 N at 0oand 400 N at 90o pull on an object. Answer the following:

a. Draw a diagram showing the forces acting on the object.b. Draw a sketch showing the vector sum of two forces.c. Find the magnitude of the resultant force.d. Find the direction of the resultant force.


a. Draw a diagram showing the forces acting on the object.


b. Draw a sketch showing the vector sum of two forces.


c. Find the magnitude of the resultant force.


d. Find the direction of the resultant force.

8. A ship makes three displacements in the following order:1) 76 mi, 48o north of east2) 50 mi, 56o north of west; and3) 47 mi, south

a. Draw a clear diagram showing all three displacement vectors with respect to horizontal points (north, east, south, and west).

b. Find the X and Y components of displacement D1.c. Find the X and Y components of displacement D2.d. Find the X and Y components of displacement D3.e. Find the magnitude of the resultant vector.f. Find the direction of the resultant vector.

X (mi)

Y (mi)

D1D2 D3 Σ




b. Find the X and Y components of displacement D1.

X (mi)

Y (mi)

D1D2 D3 Σ


c. Find the X and Y components of displacement D2.

X (mi)

Y (mi)

D1D2 D3 Σ


d. Find the X and Y components of displacement D3.

X (mi)

Y (mi)

D1D2 D3 Σ


e. Find the magnitude of the resultant vector.

X (mi)

Y (mi)

D1D2 D3 Σ


f. Find the direction of the resultant vector.

X (mi)

Y (mi)

D1D2 D3 Σ

9. A bus makes three displacements in the following order:1) 58 mi, 38o north of east2) 69 mi, 46o north of west; and3) 75 mi, southeast


b. Find the X and Y components of displacement D1.c. Find the X and Y components of displacement D2.d. Find the X and Y components of displacement D3.e. Find the magnitude of the resultant vector.f. Find the direction of the resultant vector.

X (mi)

Y (mi)

D1D2 D3 Σ




b. Find the X and Y components of displacement D1.

X (mi)

Y (mi)

D1D2 D3 Σ


c. Find the X and Y components of displacement D2.

X (mi)

Y (mi)

D1D2 D3 Σ


d. Find the X and Y components of displacement D3.

X (mi)

Y (mi)

D1D2 D3 Σ


e. Find the magnitude of the resultant vector.

X (mi)

Y (mi)

D1D2 D3 Σ


f. Find the direction of the resultant vector.

X (mi)

Y (mi)

D1D2 D3 Σ

10. A ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. How much later does the ball hit the ground?b. How far from the building will it land?c. What is the velocity of the ball just before it hits the ground?


a. How much later does the ball hit the ground?


b. How far from the building will it land?


c. What is the velocity of the ball just before it hits the ground?

11. A baseball is hit with an initial speed of 50 m/s at an angle of 30o above the horizontal.


projectile.d. Determine the velocity of the projectile 5 s after firing.










d. Determine the velocity of the projectile 5 s after firing.



projectile.d. Determine the velocity of the projectile just before it hits the

bottom of the cliff.








d. Determine the velocity of the projectile just before it hits the bottom of the cliff.

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