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

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google