An Analysis of Projectile Motion Projectile Motion Lab Report
Projectile motion (2)
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Transcript of Projectile motion (2)
PROJECTILE MOTION
Introduction
Projectile Motion:
Motion through the air without a propulsion Examples:
Part 1.Motion of Objects Projected
Horizontally
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
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?
x
y
0
Frame of reference:
h
v0
Equations of motion:
y = h + ½ g t2x = v0 tDSPL.
vy = g tvx = v0VELC.
ay = g = -9.81 m/s2
ax = 0ACCL.
Y
Accel. m.
X
Uniform m.
g
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
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
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
VELOCITY
v
vx = v0
vy = g tv = √vx
2 + vy2
= √v02+g2t2
tg Θ = vy
/ vx
= g t / v0
Θ
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)
HORIZONTAL THROW - Summary
v = √ v02+ 2h(-g)
tg Θ = -√2h(-g) / v0
Final Velocity
Δx = v0 √ 2h/(-g)Horizontal Range
Δt = √ 2h/(-g)Total time
Half -parabola, open down
Trajectory
h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2
Part 2.Motion of objects projected at an
angle
vi
x
y
θ
vix
viy
Initial velocity: vi = vi [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:
y = v0 t sin Θ + ½ g t2
x = v0x t = v0 t cos Θ
x = v0 t cos Θ
DISPLACEMENT
vy = v0 sin Θ - g t
vx = v0x= v0 cos ΘVELOCITY
ay = g = -9.81 m/s2ax = 0ACCELERATION
Y
Accelerated motion
X
Uniform motion
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
Θ+Θ=
Θ+
ΘΘ=
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 = v0 t 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 = v0 t cos Θ
Δx = vo Δt cos Θ
x
Δx
y
0
Δt =2 v0sin Θ
(-g)
Δx =2vo
2 sin Θ cos Θ
(-g)Δx =
vo 2 sin (2 Θ)(-g)
sin (2 Θ) = 2 sin Θ cos Θ
Horizontal Range, Δx
Δx =v0
2 sin (2 Θ)(-g)
0.5075
0.000
090
0.8760
1.0045
0.8730
0.5015
sin (2 Θ)Θ (deg) •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
222 cos2
tan xv
gxy
i Θ+Θ=
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
Velocity
•Final speed = initial speed (conservation of energy)
•Impact angle = - launch angle (symmetry of parabola)
Maximum Height
vy = vosin Θ + g t
y = v0 t sin Θ + ½ g t2
At maximum height vy = 0
0 = vi0sin Θ + g tup
tup = v0 sin Θ
(-g)
tup = Δt/2
hmax = v0 t upsin Θ + ½ g tup2
hmax = v02
sin2 Θ/(-g) + ½ g(vi2 sin2 Θ)/g2
hmax =
vi2
sin2 Θ
2(-g)
Projectile Motion – Final Equations
hmax =Max height
Δx =Horizontal range
Δt =Total time
Parabola, open downTrajectory
(0,0) – initial position, vi = vi [Θ]– initial velocity, g = -9.81m/s2
2 v0sin Θ
(-g)
v02 sin (2 Θ)
(-g)
v02 sin2 Θ
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