Download presentation

Presentation is loading. Please wait.

Published byShaniya Snyder Modified over 2 years ago

1
P ROJECTILE M OTION

2
Projectile Motion FThe path that a moving object follows is called its trajectory. FProjectile motion involves the trajectories and velocities of any object that has been launched, shot, or thrown. FThe path that a moving object follows is called its trajectory. FProjectile motion involves the trajectories and velocities of any object that has been launched, shot, or thrown.

3
Does this represent a realistic trajectory? a)Yes. b)No. c)Maybe. a)Yes. b)No. c)Maybe.

4
Does this represent a realistic trajectory?

5
a)Yes. b)No. c)Maybe. a)Yes. b)No. c)Maybe. The coyote would not go straight horizontally, pause, and then fall straight down. You see unrealistic trajectories all the time in media. Can you think of any others?

6
R ULES There are only a few rules we have to follow: All projectiles are freefalling vertically with an acceleration of 9.8 m/s 2 downwards Horizontal motion is totally unaffected by gravity! Since there are no forces acting on it, a falling object’s horizontal velocity is constant!

7
V ISUALIZING P ROJECTILES first enter vectors focus on v x v x is constant the whole flight!

8
V ISUALIZING P ROJECTILES first enter vectors focus on v x focus on v y v y decreases as it rises! by how much per second? no v y at the top!

9
V ISUALIZING P ROJECTILES

10
X equationsY equations 0

11
H ORIZONTAL P ROJECTILES Horizontal motion is constant – there is no acceleration Only formula used in horizontal (x) direction is: v x = d x / t constant speed!

12
H ORIZONTAL P ROJECTILES Horizontal projectiles are not thrown up or down. They are moving horizontally and falling vertically The only initial velocity is in the x direction Vertical velocity (v y ) is gained by freefall v iy = 0 since v iy is in freefall, a = -9.8 m/s 2

13
HORIZONTAL PROJECTILE pg 288 1.A rock is thrown horizontally from the top of a cliff at a constant speed of 15 m/s. Calculate the horizontal & vertical positions of the rock after each second and place these positions in the table below. Assume the rock is freefalling from rest.

14
ANGLED PROJECTILE pg 289 A rock is thrown at an angle of 30° from the top of a cliff. Calculate the horizontal & vertical positions of the rock after each second and place these positions in the table below.

15
X equationsY equations

16
1.Measure his height, Find the scale of the picture. L ET ’ S ANALYZE THE JUMP d ma x

17
L ET ’ S ANALYZE THE JUMP

18
http://www.physicsclassroom.com/mme dia/vectors/pap.cfm

19
Variable Definitions: v yi initial velocity in y direction v x = constant velocity in x direction v final velocity of projectile d x horizontal range d y Height d y Maximum altitude d y Vertical displacement v yi initial velocity in y direction v x = constant velocity in x direction v final velocity of projectile d x horizontal range d y Height d y Maximum altitude d y Vertical displacement

20
Hitting a Target If the rifle is fired directly at the target in a horizontal direction, will the bullet hit the center of the target? Does the bullet fall during its flight? If the rifle is fired directly at the target in a horizontal direction, will the bullet hit the center of the target? Does the bullet fall during its flight?

21
start by drawing a picture: E XAMPLE A person decides to fire a rifle horizontally at a bull’s-eye. The speed of the bullet as it leaves the barrel of the gun is 890 m/s. He’s new to the ideas of projectile motion so doesn’t aim high and the bullet strikes the target 1.7 cm below the center of the bull’s-eye. What is the horizontal distance between the rifle and the bull’ s-eye? label the explicit givens with subscript. Ex: d x or d y

22
givens (separate by direction): E XAMPLE What is the horizontal distance between the rifle and the bull’ s-eye? unknown: XY horizontal displacement

23
which equation do we use? E XAMPLE use to find time rewrite equation for t

24
Use t and v x to solve for d x E XAMPLE

25
N ON - HORIZONTAL P ROJECTILES v x is still constant v y is still in freefall only difference with non-horizontal is… now the object begins with a vertical velocity!

26
N ON - HORIZONTAL P ROJECTILES Angled Projectiles require a little work to get useful v i v i has an x and y component need to calculate initial v x and v y

27
B REAKING UP A VECTOR every vector has 2 components to it a horizontal component a vertical component they add up to the total

28
B REAKING UP A VECTOR SOHCAHTOA hypotenuse adjacent opposite

29
B REAKING UP A VECTOR SOHCAHTOA need to find θ? hypotenuse adjacent opposite

30
N ON - HORIZONTAL P ROJECTILES need to calculate initial v x and v y

31
1. Start by drawing a picture: A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? v i = 31.5 m/s vxvx vxvx v yi 2. Find the initial velocity in the x and y directions

32
XY A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vxvx vxvx v yi Decide upon initial and final conditions v i = 31.5 m/s

33
1.Find the velocity when the ball hits the ground, then find the time it takes to get there. 1.Use the quadratic formula A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vxvx vxvx v yi There are 2 ways to solve v i = 31.5 m/s

34
Equations A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? vxvx vxvx v yi a b c Solve using Quadratic formula … or GRAPH it! v i = 31.5 m/s

35
A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. How long is the stone in flight? v o = 31.5m/s vxvx vxvx v yo When is the stone on the ground? t = 7.00s

36
XY A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What is the range of the stone? vxvx vxvx v yo Equations v i = 31.5 m/s

37
XY A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What speed does the stone hit the ground? v f = ? vxvx vxvx v yf

38
A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What speed does the stone hit the ground? vxvx vxvx v f = ? v yf

39
Givens A NGLED P ROJECTILE E XAMPLE A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s. What speed does the stone hit the ground? Equations vxvx vxvx v yf v f = ?

40
V ARIED A NGLES which projectile angle shoots highest? larger θ means faster v iy which projectile angle shoots farthest? 45° has perfect balance of fast v x and long flight time.

41
P RACTICE PG 298

43
6. When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump 1.5 m above the groun d v = 14 m/s XY 14 m/s and 18° above the horizontal.

44
When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump? XY 1. Find t which equation do we use? We could use to find time But you would have to use the quadratic formula to solve for t because there is a t and t 2

45
When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump? XY 1. Find t which equation do we use? Instead, find v yf And then find t: v yf = - 6.9 m/s t = 1.14 s

46
When he leaves the ramp he is 1.5 m above the ground, moving at 14 m/s and 18° above the horizontal. If each of his friends is 0.60 m wide, what is the maximum number of friends over which the cyclist can jump? XY 2. Find d x v yf = - 6.9 m/s t = 1.14 s 3. How many friends?

47
V ARIED A NGLES which projectile angle shoots highest? larger θ means faster v iy which projectile angle shoots farthest? 45° has perfect balance of fast v x and long flight time.

48
` ` For the following situations: State if the following are positive, negative or zero. vovaxvovax

49
3. Calculate the maximum altitude of the ball (from the floor). XY ? 0 m/s v fy v y =

50
3. Calculate the maximum altitude of the ball (from the floor). XY ? 0 m/s v fy = from #2 v y = 1. Find t which equation do we use?

51
Percent Error Formula, last pg of text *Known = your calculation

Similar presentations

Presentation is loading. Please wait....

OK

Quick Review: Four Kinematic Equations Free Fall

Quick Review: Four Kinematic Equations Free Fall

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on power supply failure alarm Ppt on dressing etiquettes pour Ppt on entrepreneurship development Ppt on levels of management Ppt on air pollution for class 11 Ppt on pirates of silicon valley Ppt on historical places in jaipur Ppt on global warming a need for change Ppt on state of indian economy Ppt on power system protection and switchgear