Post on 05-Apr-2018
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Universiti Tun Hussein OnnMalaysiaMechanical Engineering
Dynamics DDE2063
BYEn. Khairulnizam Bin Othman
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12.6 Motion of a Projectile
Projectile: any body that is given an initial
velocity and then follows a path determined
by the effects of gravitational acceleration
and air resistance.
Trajectory path followed by a projectile
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Horizontal Motion is Uniform Motion
Notice that theHorizontal motion is in no way affected by the Vertical motion.
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Projectile
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Horizontal and vertical components of velocityare independent.
Vertical velocity decreases at a constant ratedue to the influence of gravity.
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Verify this mathematically
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Horizontal Motion
Acceleration : ax= 0
Conclusion # 1: Horizontal velocity remains constant
Conclusion # 2: Equal distance covered in equal time intervals
xxcvvtavv )()(00
tvxxtatvxx xc )(2
1
)( 002
00
xxcvvssav v )()(2)( 00
2
0
2
t
xxvx
)( 0
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Vertical Motion
ac= -g = 9.81 m/s2
Conclusion # 1: Equal increments of speed gained in equal increments
of time
Distance increases in each time interval
gtvvtavvyyc )()( 00
2
00
2
00
2
1)(
2
1)( gttvyytatvyy
yc
)(20
)(2)( 022
0
2
0
2yygssav vvv
yyc
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ProjectileMotion
Assumptions:
(1) free-fall acceleration
(2) neglect air resistance
Choosing the y direction as positive upward:
ax = 0; ay = - g (a constant)
Take x0= y0 = 0 at t = 0
Initial velocity v0 makes anangle 0 with the horizontal
v0
x
y
v v v vx y0 0 0 0 0 0
cos sin
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Maximum Height
At the peak of its trajectory, vy = 0.
From
Time t1 to reach the peak
Substituting into:
g
vt
y0
1
g
vyh
y
2
2
0
max
00 gtvgtvv oyyy
2
02
1gttvy
y
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Projection Angle
The optimal angle of projection is dependent on thegoal of the activity.
For maximal height the optimal angle is 90o.
For maximal distance the optimal angle is 45o.
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10 degrees
Projection angle = 10 degrees
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10 degrees
30 degrees40 degrees45 degrees
Projection angle = 45 degrees
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10 degrees
30 degrees40 degrees45 degrees60 degrees
Projection angle = 60 degrees
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10 degrees
30 degrees40 degrees45 degrees60 degrees75 degrees
Projection angle = 75 degrees
So angle that maximizes Range(optimal) = 45 degrees
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Example
A ball is given an initial velocity of V0 = 37 m/s at an angle of = 53.1. Find
the position of the ball, and the magnitude and direction of its velocity, when t =2.00 s. Find the time when the ball reaches the highest point of its flight, andfind its height h at this point
The initial velocity of the ball has components:
v0x = v0 cos 0 = (37.0 m/s) cos 53.1 = 22.2 m/s
v0y = v0 sin 0 = (37.0 m/s) sin 53.1= 29.6 m/s
a) position
x = v0xt = (22.2 m/s)(2.00 s) = 44.4 m
y = v0yt - gt2
= (29.6 m/s)(2.00 s) (9.80 m/s2)(2.00 s)2
= 39.6 m
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Solution (con.)
Velocity
vx = v0x = 22.2 m/s
vy = v0y gt = 29.6 m/s (9.80 m/s2)(2.00 s) = 10.0 m/s
2.24450.0arctan/2.22
/0.10arctan
/3.24
)/0.10(/2.222222
smsm
sm
smsmvvvyx
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Solution (cont.)
b) Find the time when the ball reaches the highest point
of its flight, and find its height H at this point.
ssm
sm
g
vt
gtvv
y
yy
023809
629
0
2
0
1
10
./.
/.
m
ssmssm
gttvH y
7.44
)02.3)(/80.9(2
1)02.3)(/6.29(
21
22
210 1
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Solution (cont.)
c) Find the horizontal range R, (that is, the horizontal
distance from the starting point to the point at which the
ball hits the ground.)
mssmtvR x 134)04.6)(/2.22(20
)21(
210
2022220
gtvtgttvy yy
ssm
sm
g
v
tandt
y
04.6/80.9
)/6.29(22
0 20
22
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A ball traveling at 25 m/s drive off of the edge of acliff 50 m high. Where do they land?
25 m/s
Vertically
v = v0-gt
y = y0 + v0t + 1/2gt2.
v2 = v02 - 2g(y-y0).
Horizontally
x = x0 + (v0)x t x = 25 *3.19 = 79.8 m
79.8 m
Initial Conditions
vx = 25m/s
vy0 = 0m/s
a = 9.8 m/s2
t = 0
y0 = 0 m
y = 50 m
x0
=0 m
50 = 0+0+1/2(9.8)t2 t = 3.19 s
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Review
Example 12.11
Example 12.12
Example 12.13
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