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Specification Independent effect of motion in horizontal and vertical directions of a Specification Independent effect of motion in horizontal and vertical directions of a

Specification Independent effect of motion in horizontal and vertical directions of a - PowerPoint Presentation

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Specification Independent effect of motion in horizontal and vertical directions of a - PPT Presentation

Problems will be solvable using the equations of uniform acceleration   Projectile motion This section is devoted to understanding the motion of projectiles A projectile is an object upon which the only force acting is gravity ID: 927744

projectile motion vertical horizontal motion projectile horizontal vertical velocity time speed horizontally height ground object vertically cliff box direction

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Slide1

Specification

Independent effect of motion in horizontal and vertical directions of a uniform gravitational field.

Problems will be solvable using the equations of uniform acceleration.

 

Projectile motion

Slide2

This section is devoted to understanding the motion of projectiles

A projectile is an object upon which the only force acting is gravity

An object dropped from rest (or thrown down with an initial velocity) is a projectile (provided that the influence of air resistance is negligible)

An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible)And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible)

Projectile motion

Slide3

We only need to consider motion in 1-Direction: vertically

If it is dropped from rest then the initial velocity,

u = 0 ms

-1An object dropped from rest (or thrown down with an initial velocity) is a projectile (provided that the influence of air resistance is negligible)The motion is in the direction of the acceleration: a = g = +9.81 ms-2

Projectile motion

Slide4

An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible)

We only need to consider motion in 1-Direction: vertically (up and down)

At the maximum height,

smax, the velocity, v = 0 ms-1The motion is initially in the opposite direction of the acceleration: a = g = -9.81 ms-2

Projectile motion

Slide5

We only need to consider motion in 1-Direction: vertically (up and down and past the ground level)

At the maximum height,

s

max, the velocity, v = 0 ms-1The velocity,

v

, when the ball has reached the initial position,

s = 0

is the same as the initial velocity,

u

An object that is thrown vertically upward is also a projectile and drops beyond the point it was released (provided that the influence of air resistance is negligible)

The motion is initially in the opposite direction of the acceleration:

a = g = -9.81 ms

-2

Projectile motion

Slide6

Problem:

A man stands on the edge of the cliff and throws a stone vertically upwards at 15ms

-1

. After what time will the stone hit the ground 20m below?a = g = -9.81 ms-2

At the maximum height,

s

max

, the velocity,

v = 0 ms

-1

The answer is ~4s, you work it out!

Option 1:

Calculate the time and distance for the journey to the top,

Assume it is being dropped from the top at rest with a new value for s: distance to the top + 20m, and calculate the time

Add both times together

Option 2:

Calculate the time to travel all the way up and back to its original height, so s = 0m

Calculate the time from ground level to 20m assuming it was dropped with initial speed the same as the original speed

Projectile motion

Slide7

A

B

A golfer hit ball A. It took 5 seconds to land in the position shown.

He then hit ball B with less force. How long did it take to land in its position?

4 seconds

5 seconds

6 seconds

Projectile motion

Slide8

Projectiles in 2 directions

Horizontally and vertically

Projectile motion

Slide9

Package drop

The package follows a parabolic path and remains directly below the plane at all timesThe vertical velocity changes (faster, faster)

The horizontal velocity is constant!

Slide10

The vertical velocity changes (faster, faster)

The horizontal velocity is constant!

Projectile motion

Slide11

The time taken for the object to reach the ground along the parabolic path is the same as if it were dropped vertically.

 

The parabolic path for an object projected horizontally is shown in the diagram  Notice that if air resistance is ignored the vertical height of the object at given times after the start is the same no matter what horizontal velocity it had at the moment of release.

Projectile motion

Slide12

A parcel is released from an aircraft travelling horizontally at a speed of 120ms

-1

above level ground.

The parcel hits the ground 8.5 seconds later. Calculate:The height of the aircraft above the groundThe horizontal distance travelled in this time by i) the parcel, ii) the aircraftThe speed of impact of the parcel on the ground.

Taking vertical motion

s =

½

gt

2

s = ½ x 9.8 x 8.5

2

= 354 m

b. Taking horizontal motion s = v x t

s = 120 x 8.5 = 1020m

This is the distance travelled by the aircraft and the parcel

c. Taking vertical motion g = v/t

vertical speed v = 9.8 x 8.5 = 83.3 ms

-1

Horizontal speed = 120 ms

-1

(resultant speed)

2

= 83.3

2

+ 120

2

Resultant speed = 146 ms

-1

Slide13

2. An object is projected horizontally at a speed of 16ms

-1

into the sea from a cliff top of height 45.0m. Calculate:

How long it takes to reach the seaHow far it travels horizontallyIts impact vertical velocity.Taking vertical motions=

½

gt

2

45 = ½ x 9.8 x t

2

t = 3 s

45.0m

b. Taking horizontal motion

s = v x t

s = 16 x 3 = 48m

c.

The vertical motion

g = v/t v =

gt

v = 9.8 x 3 = 29.4 ms

-1

16ms

-1

Slide14

3. A dart is thrown horizontally along a line passing through the centre of a dartboard, 2.3 m away from the point at which the dart was released. The dart hits the dartboard at a point 0.19m below the centre. Calculate:

The time of flight of the dart

Its horizontal speed of projection.

2.3 m

Taking vertical motion

s=

½

gt

2

0.19 = ½ x 9.8 x t

2

t = 0.197 s = 0.2 s

Taking horizontal motion

s = v x t v = s/t

v = 2.3/0.197 = 11.7 ms

-1

Slide15

Physics Support MaterialsHigher

Mechanics and Properties of MatterA box is released from a plane travelling with a horizontal velocity of 300 m s

-1 and at a height of 300 m. Find:

43

Click the mouse to continue

Equations of Motion

a) how long it takes the box to hit the ground

Consider the vertical motion of the box

Down is the positive direction

Slide16

Physics Support MaterialsHigher

Mechanics and Properties of MatterA box is released from a plane travelling with a horizontal velocity of 300 m s

-1 and at a height of 300 m. Find:

43 continued

Click the mouse to continue

Equations of Motion

b) the horizontal distance between the point of impact and the release point

Consider the horizontal motion of the box

c) the position of the plane relative to the box at the time of impact

The box and the plane have the same horizontal motion (constant speed), so they travel the distance horizontally as the box falls. At impact, the plane is vertically above the box.

Slide17

Physics Support MaterialsHigher

Mechanics and Properties of MatterA projectile is fired horizontally from the edge of a cliff at 12 m s

-1 and hits the sea 60 m away. Find:

44

Click the mouse to continue

Equations of Motion

the time of flight

the height of the starting point above the sea level

Consider the horizontal motion of the projectile

60 m

Sea

12 m s

-1

Slide18

Physics Support MaterialsHigher

Mechanics and Properties of MatterA projectile is fired horizontally from the edge of a cliff at 12 m s

-1 and hits the sea 60 m away. Find:

44 continued

Click the mouse to continue

Equations of Motion

b) the height of the starting point above the sea level

60 m

Sea

12 m s

-1

Consider the vertical motion of the projectile

Down is the positive direction

Slide19

Physics Support MaterialsHigher

Mechanics and Properties of MatterA ball is projected horizontally at 15 m s

-1 from the top of a vertical cliff. It reaches the horizontal ground 45 m from the foot of the cliff.

45

Click the mouse to continue

Equations of Motion

a) Draw graphs, giving appropriate numerical values of the ball’s

i

) horizontal speed against time

5

10

0

1

3

v / m s

-1

t / s

15

Calculating the time of flight:

Consider the horizontal motion of the projectile

2

Slide20

Physics Support MaterialsHigher

Mechanics and Properties of MatterA ball is projected horizontally at 15 m s

-1 from the top of a vertical cliff. It reaches the horizontal ground 45 m from the foot of the cliff.

45 continued

Click the mouse to continue

Equations of Motion

a) Draw graphs, giving appropriate numerical values of the ball’s

ii) vertical speed against time

10

20

0

1

3

v / m s

-1

t / s

30

Calculating the final vertical velocity:

Consider the vertical motion of the projectile

2

Slide21

Physics Support MaterialsHigher

Mechanics and Properties of MatterA ball is projected horizontally at 15 m s

-1 from the top of a vertical cliff. It reaches the horizontal ground 45 m from the foot of the cliff.

45 continued

Click the mouse to continue

Equations of Motion

b) Use a vector diagram to find the velocity of the ball 2 s after its projection.

After 2 s the horizontal velocity is 15 m s

-1

After 2 s the vertical velocity is 20 m s

-1

15

m s

-1

20

m s

-1

v

v

2

= 15

2

+ 20

2

v

2

= 625

v = 25

m s

-1