PROJECTILE MOTION Senior High School Physics

1 / 32

# PROJECTILE MOTION Senior High School Physics - PowerPoint PPT Presentation

PROJECTILE MOTION Senior High School Physics. Mr. T. P. Eagan 2012. Part 1. Part 2. Free powerpoints at http://www.worldofteaching.com. Introduction. Projectile Motion: Motion through the air without a propulsion Examples:.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PROJECTILE MOTION Senior High School Physics

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### PROJECTILE MOTIONSenior High School Physics

Mr. T. P. Eagan

2012

Part 1.

Part 2.

Free powerpoints at http://www.worldofteaching.com

Introduction
• Projectile Motion:

Motion through the air without a propulsion

• Examples:

Part 1.Motion of Objects Projected HorizontallyThis presentation is dedicated to all of David’s pencil’s that have gone misplaced for the last four years (in class)!!! In no way would I ever dedicate anything to Beeton!!!!

y

v0

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

x

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 final velocity?

y

v0

g

h

x

0

Frame of reference:

Equations of motion:

y

h

v02 > v01

v01

x

Trajectory

x = v0 t

y = h + ½ g t2

Parabola, open down

Eliminate time, t

t = x/v0

y = h + ½ g (x/v0)2

y = h + ½ (g/v02) x2

y = ½ (g/v02) x2 + h

y

h

Δt = √ 2h/(9.81ms-2)

Δt = √ 2h/(-g)

x

Total Time, Δt

Δt = tf - ti

y = h + ½ g t2

final y = 0

0 = h + ½ g (Δt)2

ti =0

Solve for Δt:

tf =Δt

Total time of motion depends only on the initial height, h

y

h

Δt = √ 2h/(-g)

Δx = v0 √ 2h/(-g)

Δx

x

Horizontal Range, Δx

x = v0 t

final y = 0, time is the total time Δt

Δx = v0 Δt

Horizontal range depends on the initial height, h, and the initial velocity, v0

v

v = √vx2 + vy2

= √v02+g2t2

tg Θ = vy/ vx = g t / v0

VELOCITY

vx = v0

Θ

vy = g t

Δt = √ 2h/(-g)

tg Θ = g Δt / v0

= -(-g)√2h/(-g) / v0

= -√2h(-g) / v0

v

v = √vx2 + vy2

v = √v02+g2(2h /(-g))

v = √ v02+ 2h(-g)

FINAL VELOCITY

vx = v0

Θ

vy = g t

Θ is negative (below the horizontal line)

HORIZONTAL THROW - Summary

h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2

### Part 2.Motion of objects projected at an angle

y

Initial velocity: vi = vi [Θ]

Velocity components:

x- direction : vix = vi cos Θ

y- direction : viy = vi sin Θ

viy

vi

θ

x

vix

Initial position: x = 0, y = 0

y

a = g =

- 9.81m/s2

• 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

x

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?
Trajectory

x = vi t cos Θ

y = vi tsin Θ + ½ g t2

Parabola, open down

y

Eliminate time, t

t = x/(vi cos Θ)

y = bx + ax2

x

0 =vi sin Θ + ½ g Δt

Δt =

2 vi sinΘ

(-g)

Total Time, Δt

y = vi tsin Θ + ½ g t2

final height y = 0, after time interval Δt

0 =vi Δtsin Θ + ½ g (Δt)2

Solve for Δt:

x

t = 0 Δt

2 vi sinΘ

Δt =

(-g)

Δx =

2vi 2 sin Θ cos Θ

vi 2 sin (2Θ)

Δx =

(-g)

(-g)

Horizontal Range, Δx

x = vi t cos Θ

y

final y = 0, time is the total time Δt

Δx = vi Δt cos Θ

x

0

sin (2Θ) = 2 sin Θ cos Θ

Δx

vi 2 sin (2Θ)

Δx =

(-g)

Horizontal Range, Δx
• CONCLUSIONS:
• Horizontal range is greatest for the throw angle of 450
• Horizontal ranges are the same for angles Θ and (900 – Θ)
Velocity
• Final speed = initial speed (conservation of energy)
• Impact angle = - launch angle (symmetry of parabola)

tup=

hmax =vi t upsin Θ + ½ g tup2

hmax =vi2sin2Θ/(-g) + ½ g(vi2sin2Θ)/g2

hmax =

vi sin Θ

(-g)

vi2sin2Θ

2(-g)

Maximum Height

vy = vi sin Θ + g t

y = vi tsin Θ + ½ g t2

At maximum height vy = 0

0 = vi sin Θ + g tup

tup= Δt/2

vi 2 sin (2Θ)

(-g)

Projectile Motion – Final Equations

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

2 vi sinΘ

(-g)

vi2sin2Θ

2(-g)

PROJECTILE MOTION - SUMMARY
• Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration
• The projectile moves along a parabola