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Motion in Two Dimensions

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Choosing Coordinates & Strategy For projectile motion: ... The Trajectory of a Projectile What does the free-body diagram look like for force? – PowerPoint PPT presentation

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Title: Motion in Two Dimensions


1
Motionin Two Dimensions
  • Chapter 7.2

2
Homework
  • P. 158
  • 9,10,11
  • P.160.
  • 12,13.

3
Projectile Motion
  • What is the path of a projectile as it moves
    through the air?
  • Parabolic?
  • Straight up and down?
  • Yes, both are possible.
  • What forces act on projectiles?
  • Only gravity, which acts only in the negative
    y-direction.
  • Air resistance is ignored in projectile motion.

4
Choosing Coordinates Strategy
  • For projectile motion
  • Choose the y-axis for vertical motion where
    gravity is a factor.
  • Choose the x-axis for horizontal motion. Since
    there are no forces acting in this direction (of
    course we will neglect friction due to air
    resistance), the speed will be constant (a 0).
  • Analyze motion along the y-axis separate from the
    x-axis.
  • If you solve for time in one direction, you
    automatically solve for time in the other
    direction.

5
The Trajectory of a Projectile
  • What does the free-body diagram look like for
    force?

6
The Vectors of Projectile Motion
  • What vectors exist in projectile motion?
  • Velocity in both the x and y directions.
  • Acceleration in the y direction only.

vx (constant)
ax 0
vy (Increasing)
Trajectory or Path
  • Why is the velocity constant in the x-direction?
  • No force acting on it.
  • Why does the velocity increase in the
    y-direction?
  • Gravity.

7
Ex. 1 Launching a Projectile Horizontally
  • A cannonball is shot horizontally off a cliff
    with an initial velocity of 30 m/s. If the
    height of the cliff is 50 m
  • How far from the base of the cliff does the
    cannonball hit the ground?
  • With what speed does the cannonball hit the
    ground?

8
Diagram the problem
vi
Fg Fnet
9
State the Known Unknown
  • Known
  • xi 0
  • vix 30 m/s
  • yi 0
  • viy 0 m/s
  • a -g
  • y -50 m
  • Unknown
  • x at y -50 m
  • vf ?

10
Perform Calculations (y)
  • y-direction
  • vy -gt
  • y viyt ½ gt2
  • Using the first formula above
  • vy (-9.8 m/s2)(3.2 s) 31 m/s

11
Perform Calculations (x)
  • x-Direction
  • x vixt
  • x (30 m/s)(3.2 s) 96 m from the base.
  • Using the Pythagorean Theorem
  • v vx2 vy2
  • v (30 m/s)2 (31 m/s)2 43 m/s

12
Ex. 2 Projectile Motion above the Horizontal
  • A ball is thrown from the top of the Science Wing
    with a velocity of 15 m/s at an angle of 50
    degrees above the horizontal.
  • What are the x and y components of the initial
    velocity?
  • What is the balls maximum height?
  • If the height of the Science wing is 12 m, where
    will the ball land?

13
Diagram the problem
Fg Fnet
14
State the Known Unknown
  • Known
  • xi 0
  • yi 12 m
  • vi 15 m/s
  • ? 50
  • a -g
  • Unknown
  • ymax ?
  • t ?
  • x ?
  • viy ?
  • vix ?

15
Perform the Calculations (ymax)
  • y-direction
  • Initial velocity viy visin?
  • viy (15 m/s)(sin 50)
  • viy 11.5 m/s
  • Time when vfy 0 m/s vfy viy gt
  • t viy / g
  • t (11.5 m/s)/(9.81 m/s2)
  • t 1.17 s
  • Determine the maximum height ymax yi viyt ½
    gt2
  • ymax 12 m (11.5 m/s)(1.17 s) ½ (9.81
    m/s2)(1.17 s)2
  • ymax 18.7 m

16
Perform the Calculations (t)
  • Since the ball will accelerate due to gravity
    over the distance it is falling back to the
    ground, the time for this segment can be
    determined as follows
  • Time when ball hits the ground ymax viyt ½
    gt2
  • Since yi can be set to zero as can viy,
  • t 2ymax/g
  • t 2(18.7 m)/ (9.81 m/s2)
  • t 1.95 s
  • By adding the time it takes the ball to reach its
    maximum height to the time it it takes to reach
    the ground will give you the total time.
  • ttotal 1.17 s 1.95 s 3.12 s

17
Perform the Calculations (x)
  • x-direction
  • Initial velocity vix vicos?
  • vix (15 m/s)(cos 50)
  • vix 9.64 m/s
  • Determine the total distance x vixt
  • x (9.64 m/s)(3.12 s)
  • x 30.1 m

18
Analyzing Motion in the x and y directions
independently.
  • x-direction
  • dx vix t vfxt
  • vix vi?cos?
  • y-direction
  • dy ½ (vi vf) t vavg t
  • vf viy gt
  • dy viy t ½ g(t)2
  • vfy2 viy2 2gd
  • viy vi?sin?

19
Key Ideas
  • Projectile Motion
  • Gravity is the only force acting on a projectile.
  • Choose a coordinate axis that where the
    x-direction is along the horizontal and the
    y-direction is vertical.
  • Solve the x and y components separately.
  • If time is found for one dimension, it is also
    known for the other dimension.

20
There are only so many types of problems
  • Vix constant
  • Viy 0
  • Viy gt0 (shooting upwards at an angle)
  • Viy lt 0 (shotting down at an angle)
  • a g
  • a ltgtg
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