PROJECTILE MOTION 12Opt

PROJECTILE MOTION 12Opt - PROJECTILE MOTION Senior High...

Info iconThis preview shows pages 1–14. Sign up to view the full content.

View Full Document Right Arrow Icon
PROJECTILE MOTION Senior High School Physics Lech Jedral 2006 Part 1. Part 2. 100’s of free ppt’s from www.pptpoint.com library
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Introduction Projectile Motion: Motion through the air without a propulsion Examples:
Background image of page 2
Part 1. Motion of Objects Projected Horizontally
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
v 0 x y
Background image of page 4
x y
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
x y
Background image of page 6
x y
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
x y
Background image of page 8
x y Motion is accelerated Acceleration is constant, and downward a = g = -9.81m/s 2 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/s 2
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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?
Background image of page 10
x y 0 Frame of reference: h v 0 Equations of motion: X Uniform m. Y Accel. m. ACCL. a x = 0 a y = g = -9.81 m/s 2 VELC. v x = v 0 v y = g t DSPL. x = v 0 t y = h + ½ g t 2 g
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Trajectory x = v 0 t y = h + ½ g t 2 Eliminate time, t t = x/v 0 y = h + ½ g (x/v 0 ) 2 y = h + ½ (g/v 0 2 ) x 2 y = ½ (g/v 0 2 ) x 2 + h y x h Parabola, open down v 01 v 02 > v 01
Background image of page 12
y = h + ½ g t 2 final y = 0 y x h t i =0 t f =Δt 0 = h + ½ g (Δt) 2 Solve for Δt:
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 14
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 32

PROJECTILE MOTION 12Opt - PROJECTILE MOTION Senior High...

This preview shows document pages 1 - 14. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online