PROJECTILE MOTION Senior High School Physics
-
Upload
phelan-olson -
Category
Documents
-
view
58 -
download
0
description
Transcript of PROJECTILE MOTION Senior High School Physics
![Page 1: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/1.jpg)
PROJECTILE MOTIONSenior High School Physics
Lech Jedral
2006
Part 1. Part 2.
Free powerpoints at http://www.worldofteaching.com
![Page 2: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/2.jpg)
Introduction
Projectile Motion:
Motion through the air without a propulsion Examples:
![Page 3: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/3.jpg)
Part 1.Motion of Objects Projected
Horizontally
![Page 4: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/4.jpg)
v0
x
y
![Page 5: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/5.jpg)
x
y
![Page 6: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/6.jpg)
x
y
![Page 7: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/7.jpg)
x
y
![Page 8: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/8.jpg)
x
y
![Page 9: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/9.jpg)
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
![Page 10: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/10.jpg)
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?
![Page 11: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/11.jpg)
x
y
0
Frame of reference:
h
v0
Equations of motion:
X
Uniform m.
Y
Accel. m.
ACCL. ax = 0 ay = g = -9.81 m/s2
VELC. vx = v0 vy = g t
DSPL. x = v0 t y = h + ½ g t2
g
![Page 12: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/12.jpg)
Trajectory
x = v0 t
y = h + ½ g t2
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
x
hParabola, open down
v01v02 > v01
![Page 13: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/13.jpg)
Total Time, Δty = h + ½ g t2
final y = 0 y
x
hti =0
tf =Δt
0 = h + ½ g (Δt)2
Solve for Δt:
Δt = √ 2h/(-g)
Δt = √ 2h/(9.81ms-2)
Total time of motion depends only on the initial height, h
Δt = tf - ti
![Page 14: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/14.jpg)
Horizontal Range, Δx
final y = 0, time is the total time Δt
y
x
h
Δt = √ 2h/(-g)
Δx = v0 √ 2h/(-g)Horizontal range depends on the initial height, h, and the initial velocity, v0
Δx
x = v0 t
Δx = v0 Δt
![Page 15: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/15.jpg)
VELOCITY
v
vx = v0
vy = g tv = √vx
2 + vy2
= √v02+g2t2
tg Θ = vy/ v
x = g t / v
0
Θ
![Page 16: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/16.jpg)
FINAL VELOCITY
v
vx = v0
vy = g tv = √vx
2 + vy2
v = √v02+g2(2h /(-g))
v = √ v02+ 2h(-g)
Θtg Θ = g Δt / v0
= -(-g)√2h/(-g) / v0
= -√2h(-g) / v0
Θ is negative (below the horizontal line)
Δt = √ 2h/(-g)
![Page 17: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/17.jpg)
HORIZONTAL THROW - Summary
Trajectory Half -parabola, open down
Total time Δt = √ 2h/(-g)
Horizontal Range Δx = v0 √ 2h/(-g)
Final Velocity v = √ v02+ 2h(-g)
tg Θ = -√2h(-g) / v0
h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2
![Page 18: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/18.jpg)
Part 2.Motion of objects projected at an
angle
![Page 19: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/19.jpg)
vi
x
y
θ
vix
viy
Initial velocity: vi = vi [Θ]
Velocity components:
x- direction : vix = vi cos Θ
y- direction : viy = vi sin Θ
Initial position: x = 0, y = 0
![Page 20: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/20.jpg)
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
![Page 21: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/21.jpg)
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?
![Page 22: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/22.jpg)
Equations of motion:
X
Uniform motion
Y
Accelerated motionACCELERATION ax = 0 ay = g = -9.81 m/s2
VELOCITY vx = vix= vi cos Θ
vx = vi cos Θ
vy = viy+ g t
vy = vi sin Θ + g tDISPLACEMENT x = vix t = vi t cos Θ
x = vi t cos Θ
y = h + viy t + ½ g t2
y = vi t sin Θ + ½ g t2
![Page 23: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/23.jpg)
Equations of motion:
X
Uniform motion
Y
Accelerated motionACCELERATION ax = 0 ay = g = -9.81 m/s2
VELOCITY vx = vi cos Θ vy = vi sin Θ + g t
DISPLACEMENT x = vi t cos Θ y = vi t sin Θ + ½ g t2
![Page 24: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/24.jpg)
Trajectoryx = vi t cos Θ
y = vi t sin Θ + ½ g t2
Eliminate time, t
t = x/(vi cos Θ)
y
x
Parabola, open down
222
22
2
cos2tan
cos2cos
sin
xv
gxy
v
gx
v
xvy
i
ii
i
y = bx + ax2
![Page 25: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/25.jpg)
Total Time, Δt
final height y = 0, after time interval Δt
0 = vi Δt sin Θ + ½ g (Δt)2
Solve for Δt:
y = vi t sin Θ + ½ g t2
0 = vi sin Θ + ½ g Δt
Δt = 2 vi sin Θ
(-g)t = 0 Δt
x
![Page 26: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/26.jpg)
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
![Page 27: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/27.jpg)
Horizontal Range, Δx
Δx =vi
2 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)
![Page 28: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/28.jpg)
Trajectory and horizontal range2
22 cos2tan x
v
gxy
i
0
5
10
15
20
25
30
35
0 20 40 60 80
15 deg
30 deg
45 deg
60 deg
75 deg
vi = 25 m/s
![Page 29: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/29.jpg)
Velocity
•Final speed = initial speed (conservation of energy)
•Impact angle = - launch angle (symmetry of parabola)
![Page 30: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/30.jpg)
Maximum Height
vy = vi sin Θ + g t
y = vi t sin Θ + ½ g t2
At maximum height vy = 0
0 = vi sin Θ + g tup
tup = vi sin Θ
(-g)
tup = Δt/2
hmax = vi t upsin Θ + ½ g tup2
hmax = vi2
sin2 Θ/(-g) + ½ g(vi2
sin2 Θ)/g2
hmax =
vi2
sin2 Θ
2(-g)
![Page 31: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/31.jpg)
Projectile Motion – Final Equations
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 vi sin Θ
(-g)
vi 2 sin (2 Θ)
(-g)
vi2
sin2 Θ
2(-g)
![Page 32: PROJECTILE MOTION Senior High School Physics](https://reader030.fdocuments.us/reader030/viewer/2022032605/56812c44550346895d90c876/html5/thumbnails/32.jpg)
PROJECTILE MOTION - SUMMARY Projectile motion is motion with a constant
horizontal velocity combined with a constant vertical acceleration
The projectile moves along a parabola