2D  mo&on  -­‐PROJECTILE  MOTION  Chapter  4  

Let’s  talk  about  Unit  1  test  

•  Test  Correc&ons  acceptable  before  end  of  Monday  10/17.  Anyone  can  get  up  to  ½  the  points  lost  .  

•  You  will  write:  1.  What  mistakes  you  made  and  why?  2.  How  you  corrected  them  showing  all  steps.    3.  Then  you  will  meet  with  me  to  explain.  

Introduc&on  

•  Projec&le  Mo&on:  Mo&on  through  the  air  without  a  propulsion  •  Examples:  

Claim-­‐  Evidence  -­‐Reasoning  

Step 1 - In groups of 2 Claim something about the time of flight in the 2 cases you observe Step 2 – What evidence you have to support your Claim?

Step 3 – Reason why your evidence supports your claim. Hint: What remains same in the 2 cases ?

Observe  what  happens  to  Vy    as  mo&on  progresses  

Observe  Vx  now  

v0

x

y

x

y

x

y

x

y

x

y

x

y

• Motion is accelerated

• Acceleration is constant, and downward

•  a = -g = -9.81m/s2

• The horizontal (x) component of velocity is constant

• The horizontal and vertical motions are independent of each other, but they have a

common time

-g = -9.81m/s2

vi

x

y

θ

v0x

v0y

Initial velocity: v0 = v0 [Θ]

Velocity components:

x- direction : v0x = v0 cos Θ

y- direction : v0y = v0 sin Θ

Initial position: x = 0, y = 0

x

y

•  Motion is accelerated

•  Acceleration is constant, and downward

•  a = -g = -9.81m/s2

•  The horizontal (x) component of velocity is constant

•  The horizontal and vertical motions are independent of each other, but they have a common time

a = g =

- 9.81m/s2

ANALYSIS OF MOTION: ASSUMPTIONS

•  x-direction (horizontal): uniform motion

•  y-direction (vertical): accelerated motion

•  no air resistance

QUESTIONS

•  What is the trajectory?

•  What is the total time of the motion?

•  What is the horizontal range?

•  What is the maximum height?

•  What is the final velocity?

Equations of motion:

X Uniform motion

Y Accelerated motion

ACCELERATION

VELOCITY

DISPLACEMENT

Trajectory  

x = v0 cos Θ t y = v0 sin Θ t - ½ g t2 Eliminate time, t

t = x/(v0 cos Θ)

y

x

Parabola, open down

y = v0xsinΘv0 cosΘ

−gx2

2v02 cos2Θ

y = x tanΘ− g2v0

2 cos2Θx2

y = bx + ax2

Total  Time,  Δt  

final height y = 0, after time interval Δt

0 = v0 Δt sin Θ - ½ g (Δt)2

Solve for Δt:

y = v0t sin Θ - ½ g t2

0 = v0 sin Θ - ½ g Δt

Δt = 2 v0 sin Θ

g t = 0 Δt

x

Horizontal  Range,  Δx  

final y = 0, time is the total time Δt

x = vi t cos Θ

Δx = vi Δt cos Θ

x

Δx

y

0 Δt =

2 vi sin Θ

(-g)

Δx =

2vi 2 sin Θ cos Θ (-g)

Δx = vi 2 sin (2 Θ) (-g)

sin (2 Θ) = 2 sin Θ cos Θ

Horizontal  Range,  Δx  

Δx = v02 sin (2 Θ) g

Θ (deg) sin (2 Θ) 0 0.00

15 0.50

30 0.87

45 1.00

60 0.87

75 0.50

90 0

• CONCLUSIONS:

• Horizontal range is greatest for the throw angle of 450

•  Horizontal ranges are the same for angles Θ and (900 – Θ)

Trajectory  and  horizontal  range  

y = x tanΘ− g2v0

2 cos2Θx2

0

5

10

15

20

25

30

35

0 20 40 60 80

15 deg30 deg45 deg60 deg75 deg

v0= 25 m/s

Velocity  

• Final speed = initial speed (conservation of energy)

• Impact angle = - launch angle (symmetry of parabola)

Cannonball to pirate ship Figure shows a pirate ship 560 m from a fort defending a harbor entrance. A defense cannon, located at sea level, fires balls at initial speed . At what angle from the horizontal must a ball be fired to hit the ship? What is the Max Range of the Cannon ball?

Projectile dropped from airplane In the Fig, a rescue plane flies at 198 km/h (= 55.0 m/s) and constant height toward a point directly over a victim, where a rescue capsule is to land. 1.  What should be the angle of the pilot's line of sight to the victim when the capsule release is made?

2. As the capsule reaches the water, what is its velocity in unit-vector notation and in magnitude-angle notation?

Projec&le  Mo&on  –  Final  Equa&ons  

Trajectory Parabola, open down

Total time

Δt =

Horizontal range

Δx =

Max height

hmax =

(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2

2 v0 sin Θ g

v0 2 sin (2 Θ)

g

v02 sin2 Θ

2g

PROJECTILE  MOTION  -­‐  SUMMARY  

•  Projec&le  mo&on  is  mo&on  with  a  constant  horizontal  velocity  combined  with  a  constant  ver&cal  accelera&on    

•  The  projec&le  moves  along  a  parabola