Year 12 Physics Vertical Motion (part 2) 2020

12 Physics Craighead

Townshend 1

Year 12 Physics

Vertical Motion

(part 2)

2020

NAME: _______________________


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Solving simple physics problems

• Can identify situation when an object is being affected by constant acceleration in a straight line.

• Can use the equation to work out the final velocity of a climbing or falling object.

• Can use the equation to work out the distance a climbing or falling object.


• Can rearrange any of the kinematic equations to make any of the symbols the subject.

• Can identify situations where projectile motion is involved.

• Can use atvv if += to work out the initial velocity for the vertical aspect of the projectile motion when

given the time and vice versa.

• Can use

21

2id v t at= +

to work out the range or height an object travels.

• Can use advv if 222+= to work out the initial velocity for the vertical aspect of the projectile motion

when given the height an object reaches and vice versa.

• Can use the given angle and the initial velocity to work out the initial vertical and horizontal velocities.

• Can work out the initial velocity when given the initial horizontal or vertical velocity and the angle.


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Motion under constant acceleration

• Can identify situation when an object is being affected by constant acceleration in a straight line.

When an object has a constant unbalanced force acting on it, it will be under constant acceleration.

When an object is constantly accelerating the velocity must be changing. Average velocity, v is no longer

useful and we have to use instantaneous velocities in our calculations. When we investigate the motion

of an accelerating object over a period of time, t, vi is the initial velocity of the object and vf is the final

velocity.

Quantity Symbol Unit

average velocity v m s-1

initial velocity vi m s

-1

final velocity vf m s

-1

distance d m

time t s

acceleration a m s-2

Final velocity (vf).

The final velocity of an object is an instantaneous velocity and we can work it out by adding the change

in the velocity to the initial velocity (vi).

We use this equation to work out the final velocity of an object that has undergone constant

acceleration for a known time.

Exercises

1. What is the final velocity of a car that was traveling at 20 m s-1 and accelerated constantly

at 4.0 m s-2 for 6.0 s?

11

120

4.0

6.0

44

fi

f

vv

at

vm

sm

ss

ms

−−

−

=+

=+

=

f iv v at= +


Townshend 5

Value

d subject

a -9.8 m s-2

vi 0.0 m s-1

vf unknown

t 1.25 s

2. A girl undergoes constant acceleration from 4.1 m s-1 to 8.3 m s-1 for 3.7 seconds, what is her

acceleration?

Vertical Motion

Objects that undergo vertical motion have a constant acceleration, due to gravity, acting on them (as

long as gravity is the only force that is acting on them). When we solve vertical motion problems, we

ignore the effects of friction and we assign the acceleration due to gravity, g as -9.8 m s-2. The

acceleration is negative as its direction directly down, perpendicular to the surface of the Earth. If the

displacement or velocity is also down it is negative, if it is up it is positive.

The physical quantities that we use in vertical motion calculations at vi, vf, d, a and t. The following three

equations are used to find out unknown quantities.

Example

A water balloon is dropped from the top of the piazza at Caroline Bay. The balloon took 1.25 s to hit the

ground once it was released. How high is the piazza?

subject = d

t = 1.25 s

vi = 0.0 m s-1

As vi = 0 the vit = 0

d is negative as the displacement is downwards

2 2 2f iv v ad= +21

2id v t at= +

2 1 2 2

2 2

1 10.0 1.25 9.8 (1.25 )

2 2

19.8 (1.25

) 7.66

2

id v t at ms s ms s

d ms ms

− −

−

= + = + −

= − = −

f iv v at= +


Townshend 6

Rearranging equations

Rearrange the equations so vi is the subject.

Rearrange the equations so t is the subject.

Rearrange the equations so d is the subject.

2 2 2f iv v ad= +f iv v at= + 21

2id v t at= +

f iv v at= +

2 2 2f iv v ad= +


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5 m

3 m

Aim

To find the relationship between the distance a mass falls, d and time, t. Use this to find the

acceleration due to gravity, a.

Equipment

1 metre ruler, hockey ball, school iPad.

Theory For an object that was stationary and then falls, the following equation can be used to relate

distance, acceleration and time:

Method

Video Physics

1. Set up the experiment as in the

diagram to the right.

2. Drop the hockey ball from a

height of 3 meters.

3. Record the time for each

distance.

4. Repeat the experiment two

more times

Results

1. Record all raw measurements in an appropriately headed results table.

d (m) t (s)

test 1

t (s)

test 2

t (s)

test 3 tav (s) T2 ( )

0.50 m

1.00 m

1.50 m

2.00 m

2.50 m

3.00 m

hockey ball

Reference stick

21

2id v at= +


Townshend 9

Analysis of data

2. State the type of relationship shown on the Video Physics distance/time graph between d and t.

• Transform the data to obtain a straight-line graph. Record all transformed data.

• Use the information from the straight line graph, find and state the mathematical relationship

between d and t.

Conclusion

• State your findings in relation to your AIM.


Townshend 10

Value

d

a

vi

vf

t

Value

d

a

vi

vf

t

Value

d

a

vi

vf

t

Vertical Motion


• Can use the equation to work out the distance a climbing or falling object.


Exercises

3. A.J. Hacket’s bungee platform is 43 m above the Kawarau River. Mr Townshend’s shoe fell off the

jumping platform. How long would it take the

shoe to fall to the river?

Hint: Start by making t the subject of the equation.

4. A girl kicks a ball vertically up into the air at 4.3 m s-1. Assuming that the ball is kicked from

ground level, what is the maximum height the ball reaches?

5. How long does it take for the ball to reach its highest point?


Townshend 11

Value

d subject

a -9.8 m s-2

vi 0.0 m s-1

vf unknown

t 1.25 s

Value

d

a

vi

vf

t

6. In order to break open mussel shells, sea gulls drop

them from height onto rocks. An average shell needs to

hit the rocks at 16.3 m s-1 to break. How high must the

sea gull release the shell from in order

for it to break?

7. The Eiffel tower is 324 m high. A ball is dropped from the

top of the Eiffel tower. What is the balls velocity just

before it hits the ground?

http://maineanencyclopedia.com/wpcontent/uploads/03020106-Copy.jpg


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Townshend 14

Projectile motion

• Can draw a diagram to represent the height, range and initial or final velocity.

An object undergoes projectile motion when it travels through the air with gravity being the

only force acting on it. The force of gravity, Fg, provides a constant acceleration, a of -9.8 m s-1

downwards on all objects undergoing projectile motion. It is this acceleration that produces the

characteristic parabolic path of projectiles.

8. Are the following objects undergoing projectile motion? Explain your answer.

A rocket lifting into orbit.

A javelin just after it has left the hand of an athlete.

A car that has just been driven off a cliff.

An aeroplane flying through the air.

A paper plain flying through the air.

Fg

Fg

Fg

Fg

Fg


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Velocity during projectile motion

• Identify the relative horizontal and vertical velocities of an object at different positions in its parabolic

motion.

• Can separate the quantities given into vertical or horizontal motion.

• Can separate the initial (or final) velocity into its component vertical and horizontal velocities.

At any given point the instantaneous velocity of a projectile can be represented as a

combination of horizontal and vertical velocities. The vertical and horizontal velocities are said

to be components of the overall velocity.

Vertical component of velocity

The constant acceleration from gravity acts in the opposite direction to the initial vertical

velocity causing the vertical component of velocity to decelerate. The projectile eventually

reaches its highest point on its path. At this point, the projectile’s vertical velocity equals

0.0 m s-1. After the highest point the projectile continues to accelerate downwards. The

direction of the vertical velocity will become negative as it is now moving downwards.

Horizontal component of velocity

There are no forces acting in the horizontal direction (we ignore any friction present) so the

horizontal velocity is constant throughout the duration of the projectile’s flight.

horizontal

component

vertical

component

Overall

velocity


Townshend 16

Exercises

Compare the initial vertical and horizontal velocities of A and B.

Compare the initial vertical and horizontal velocities of B and C.

Initial velocity

When we carry out calculations involving objects undergoing projectile motion we often use the

terms initial velocity and final velocity. Initial velocity is a vector giving both the magnitude and

direction (angle from the ground) of the combined horizontal and vertical velocities at the

starting point. In order to solve problems with this data we need to separate this initial velocity

into its vertical and horizontal components.

We can use the following symbols in our calculations.

vi = initial vertical velocity

vf = final vertical velocity

v = horizontal velocity (both initial and final as horizontal velocity as it is constant)

We can calculate vi and v from the initial velocity using trigonometry. (in the same way we can

calculate vf and v using trigonometry from the final velocity)

28°

v

28° vi

B C

A

1

1

sin

sin 2 8 14

6.6

i

i

i

v initialvelocity

v m s

v m s

−

−

=

=

=

1

1

cos

cos2

8 14

12

v initialvelocity

v ms

v ms

−

−

=

=

=


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Projectile motion questions

1. A ball kicked at a 30° angle with an initial velocity of 15 ms-1

a. What is the initial horizontal velocity?

b. What is the initial vertical velocity?

c. How long does it take for the ball to reach its highest point?

d. What height does the ball reach?

e. How long is the ball in the air for?

f. How far does the ball travel horizontally?

1.4

3 s


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2. A motorbike rides off a ramp with a 15° angle at 45 ms-1

a. What is the initial horizontal velocity?

b. What is the initial vertical velocity?

c. How long does it take for the motorbike to reach its highest point?

d. What height does the motorbike reach?

e. How long is the motorbike in the air for?

f. How far does the motorbike travel horizontally?


Townshend 20

Steps involved in Projectile Motion problems

Step description Equations 1 Calculate the component vertical and horizontal

velocities

2 Calculate the height when vf = 0 m s-1

3 Calculate the time when vf = -vi

4 Calculate the horizontal distance (range)

5 Calculate the time at a given final velocity.

6

Calculate vi by using vf = 0 m s-1 at half way. Time equals half the total time.

7 Calculate the vertical velocity at a given height.

8 Calculate the final velocity (or initial) at a given height by combining the vf and v vectors

9 Calculate the time, given the range

10

Use a velocity angle and component to find the combined velocity.

11 Calculate horizontal velocity, v, given range and time.

siniv initialvelocity=

21

2id v t at= +

f iv v at= +

2 2 2f iv v ad= +

f iv v at= +

cosv initialvelocity=

1tan iv

v −= 2 2c a b= +

f iv v at= +

sin

ivinitialvelocity

=


Townshend 21

Value

d unknown

a -9.8 m s-2

vi 14.44 m s-1

vf -14.44 m s-1

t subject

Clearly identify and show the equations and steps needed to answer the following questions.

3. A projectile is launched at 24 ms-1 at an angle of 37⁰. What horizontal distance did it travel?

STEP 1

STEP 3

STEP 4

1 19.17 2.947

56.40 56

dv

m m

t

d v t ms s

d

−

=

=

=

=

=

24 ms-1

37 ⁰

subject = d (range)

STEP 4

In order to find v we need to do Step 1

In order to find t we need to do Step 3

STEP 1

STEP 3

STEP 1

subject

STEP 3

STEP 4

v

37°

Keep all working to at least 4

sig. fig. until the final answer.

1

1

sin

sin 37

24

14.44

i

i

i

v initialvelocity

v m s

v m s

−

−

=

=

=

1

1

cos

cos37 28

19.17

v initialvelocity

v ms

v ms

−

−

=

=

=

f i

f i

f i

v v at

v v at

v vt

a

= +

− =

−=

1 1

2

1

2

14.44 (14.44 )

9.8

28.882.947

9.8

ms mst

ms

mst s

ms

− −

−

−

−

− −=

−

−= =

−


Townshend 22

4. A projectile is launched at an angle of 65⁰ and travels for 4.7 s until it hits the ground again. What

distance did it travel vertically and horizontally? Write a flow chart of the step you will have to carry

out to solve this problem.

5.

65 ⁰

t = 4.7 s


Townshend 23

6. A projectile is launched and travels a distance of 359 m in 13.2 s. What is the initial velocity and

angle the projectile is launched at?

d = 359 m

t = 13.2 s


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7. A projectile is launched and travels a distance of 158 m while reaching a height of 41 m. What is the

final velocity and angle the projectile hits the ground at?

h = 41 m

d = 158 m


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Key Word Definition Initial velocity

vertical

horizontal

Final velocity

projectile

gravity

height

range

parabola


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End of topic – Practice Test – Projectile motion

Complete the following question as a practice for the end of topic test.


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Feedback/feedforward


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Year 12 Physics Vertical Motion (part 2) 2020

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Transcript of Year 12 Physics Vertical Motion (part 2) 2020