Post on 13-Nov-2014
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2.4 PROJECTILE MOTION
LESSON 4
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• To describe and use equations of projectile motion,
vx=vo cos ; vy=vo sin ax=0; ay=g=-9.81m s-2
• To derive equations for maximum height, H, horizontal range, R and time of flight, T.
Objectives:
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• Projectile motion is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration
• Horizontal motion:
- velocity in the x-direction is constant ; vx = constant, ax = 0.• Vertical motion: - velocity in the y-direction has a constant acceleration of g (air resistance is
ignored) ; ay = g = -9.81 m s-2.
2.4 : Projectile Motion
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Parabola – the path is symmetrical
Where ; vo = initial velocityv = velocity at time tvy = velocity of y-component at time tvox = initial x-component velocityvoy = initial y-component velocitysx = horizontal rangeh = maximum height
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Equations of projectile motionEach vector quantity is resolved into 2 components - x component
- y component
Quantity Horizontal motion (x-component)
Vertical motion (y-component)
Displacement sx = voxt sy = voyt - ½ gt2
Velocity vx = vox (constant
velocity)
vy = voy - gt
Acceleration ax = 0 ax = g = 9.81 m s-2
(constant acceleration)
v0
v0x
v0y
x – component ; v0x = v cos
y – component ; v0y = v sin
ox
oy
v
v1tan
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Velocity at any point;
vx
vy
Equations of projectile motion
22v ; Magnitude yx vv
x
y
v
v1-tan ;Direction
v
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Maximum Range and HeightMaximum Range and Height
An object is launched at an initial velocity v0 at an angle with the horizontal.
H
Sx= R
x
y vy= 0
sy =0)
v0
Maximum Height , H
Vertically:
Initial velocity, v0y= v0 sin ,
At maximum height, vy = 0,
Using
vy2 = vo
2 + 2as
0 = (v0sin )2 + 2 (-g)H
2 20 sin
2
vH
g
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Vertically:
At the highest point, vy =0,
Using vy = voy – gt ;
0 = v0 sin - gt
The time taken to reach
the maximum height,
Time of flight, T
H
Sx= R
x
y vy= 0
sy =0)
v0
( sin )ovt
g
2 sin2 ov
T tg
As the path is symmetrical, time to go up = time to come down.
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Horizontally: ax= 0 ; vox = vo cos
sx = v0xt;
20 sin 2v
Rg
Range, R
The range is the horizontal displacement for the projectile.
R is maximum when 2 = 900
where sin 2 = 2 sin cos
90sin
2
max
2
g
vR
g
vR
o
oo
H
Sx= R
x
y vy= 0
sy =0)
v0
(2 sin )( cos ) o
o
vR v
g
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Maximum Range and Height• What are the conditions that give maximum
height and range in a projectile motion?
g
vh ii
2
sin 22 Maximum height can be achieved when i=90o!!!
2 sin 2i ivR
g
Maximum range can be achieved when 2i= 90o, i.e., i=45o!!!
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a) the maximum height ,Hb) the time of flight,c) the horizontal range, R,d) its velocity when it strikes the ground,
Example 1 – Projectile motion
A ball is launched with a velocity 20 m s-1 at an angle 60o with the horizontal at a point A. Calculate
a) H ?
c) Sx= R ?
x
y
Sy =0b) t ?
A
d) v ?
600
v0 = 20 ms-1
vh = 0
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a) H ?
c) Sx= R ?
x
y
Sy =0b) t ?
A
d) v ?
600
V0 = 20 ms-1
vh = 0
Solution of example 1 – Projectile motion;
A ball is launched with a velocity 20 m s-1 at an angle 60o with the horizontal at a point A.
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(a) Maximum height, H
Vertically: voy= 20 sin 600 ; vy = 0 (at maximum height)
Using vy2 = voy 2 – 2gs
0 = (20 sin 600)2 - 2(9.81)(H) H = 15.29 m
Solution of example 1 – Projectile motion
(b) Time of flight, T
Vertically:
voy= 20 sin 600 ; sy=0 ( Back on ground)
Using sy = v0yt - 1/2 gyt2
0 = v0 sin t - ½ gt2
Time of flight, T = 2v0 sin / g T = [2(20 sin 600 )]/9.81 = 3.53 s
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Solution of example 1 – Projectile motion
(c) Range, R
Horizontally :
vox= 20 cos 60o, T = 3.53 s, ax= 0
Using sx= voxt
R = (20 cos 60o)(3.53)
= 35.3 m
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(d) Velocity when the ball strikes the ground ;
Horizontally :
vx = vox = 20 cos 60o = 10 ms-1
Vertically :
v0y = 20 sin 60o ; T = 3.53s
Using vy = voy -- gt
vy= 20 sin 60o - (9.81)(3.53)
= - 17.31ms-1. (downward)
vx= 10 m s-1
vy = - 17.31m s-1 v
Solution of example 1 – Projectile motion
-12222 s m 2031.1710 yx vvv
o
x
y
v
v60
10
31.17tantan 11
Magnitude:
from the diagram, v = -20 m s-1 (downward)
Direction :
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At an altitude of 500 m, a bomb is released from an airplane flying horizontally with uniform velocity, 100 m s-1 to hit a target which is located 800 m away. Find the time the bomb reaches the ground.Is the mission successful?
500 m
800 m
Example 2 – Projectile motion
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x
y
vo = 100 ms-1
sy = -500 m
t ?
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Vertically:voy = 100 sin 0o = 0 ; g = 9.81m s-2 ; sy = -500 m.
sy = voyt - 1/2 gt2
-500 = 0 - 1/2 gt2
t = 10.1 s
Horizontally:vox = 100 cos00 = 100 m s-1 ; ax = 0 ; t = 10.1 s
sx = voxt - 1/2 axt2
= (100) (10.1) + 0 = 1010 m Since the target is located 800 m, hence the bomb misses the
target and so the mission is not successful.
Solution of example 2 – Projectile motion
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A firefighter, a distance d from a burning building, directs a stream of water from a fire hose at angle i above the horizontal as in the figure. If the initial speed of the stream is vi , at what height h does the water strike the building?
Example 3 – Projectile motion
Solution :
The horizontal component of displacement,
cosx ix i is d v t v t the time required to reach the building a distance d away is ;
cosi i
dt
v
At this time, the altitude of the water ; 21
2y iyv h v t gt 2
sincos 2 cosi i
i i i i
d g dv
v v
the water strikes the building at a height h above ground level of ;2
2 2tan
2 cosy ii i
gds h d
v
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Example 4 – Projectile motionA ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s-1, 40o above horizontal. How far above or below its original level will the ball strike the opposite wall ?
)40o
20 m s-1
50 m
Solution :0 -120cos 40 15.3 m soxv 0 -120sin 40 12.9 m soyv
Horizontally ; x x oxs v t v t
503.27 s
15.3t
Vertically ;
212y oys v t gt
212(12.9)(3.27) (9.81)(3.27)
10.27 m
Therefore, the ball hit at 10.27 m below the original level.
time taken to strike the opposite wall :
original level
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Example 5 – Projectile motionA projectile is fired horizontally with a speed of 30 m s-1 from the top of a cliff 80 m high. a) How long will it take to strike the level ground at the base of the cliff ? b) How far from the foot of the cliff will it strike ?
c) With what velocity will it strike ?
80 m
30 m s-1 Solution :
80 m ;ys -130 m s ;o oxv v -10 m soyv
a) Vertically ; 212y oys v t gt
21280 (0) (9.81)t t
4.04 st b) Horizontally ;
x oxs v t(30)(4.04) 121.2 mxs c) Final velocity , v ;
-130 m s ;x oxv v -10 (9.81)(4.04) 39.6 =40 m sy oyv v gt
2 2 2 2 -1Magnitude, 30 ( 40) 50 m sx yv v v
-1 040Direction , = tan 53
30
-1 0Hence, v = 50 m s ; 53 below +ve x-axis
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A boy playing football kicks the ball at an angle of 37o with respect to the horizontal, at an initial speed of 15 m s-1. Assuming that the ball moves in a vertical plane. Calculate ;
a) the time t1 after which the ball will reach the highest point of its trajectory and the height of this point above the ground
b) the time t2 taken by the ball to fall on the ground and its speed at that moment
c) the longest possible range of the ball
Example 6 – Projectile motion
Answer :
a) t1 = 0.9 s ; ymax = 4.1 m
b) t2 = 1.8 s ; v = 15 m s-1
c) xmax = 22.9 m
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Conclusion
2 20 sin
2
vH
g
20 sin 2v
Rg
• Equations of projectile motion, vx= vo cos ; vy= vo sin ax= 0; ay= g = -9.81m s-2
• Equations for maximum height, H, horizontal range, R and time of flight, T (for a parabolic curve)
g
vT o sin2
H
R
y
x