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King Fahd University ofPetroleum & Minerals
Mechanical Engineering Dynamics ME 201
BY
Dr. Meyassar N. Al-Haddad Lecture # 4
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12.6 Motion of a Projectile
Projectile: any body that is given an initialvelocity and then follows a path determined
by the effects of gravitational accelerationand air resistance.
Trajectory path followed by a projectile
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Horizontal Motion is Uniform Motion
Notice that the Horizontal motion is in no way affected by the Vertical motion .
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Verify this mathematically
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Horizontal Motion Acceleration : a x= 0
Conclusion # 1 : Horizontal velocity remains constant Conclusion # 2 : Equal distance covered in equal time intervals
x xc vvt avv )()( 00
t v x xt at v x x xc )(21
)( 002
00
x xc vv s sav v )()(2)( 002
02
t
x xv x
)( 0
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Vertical Motion
ac= -g = 9.81 m/s 2 = 32.2 ft/s 2
Conclusion # 1 : Equal increments of speed gained in equal incrementsof time
Distance increases in each time interval
gt vvt avv y yc )()( 00
200
200
2
1)(
2
1)( gt t v y yt at v y y yc
)(20)(2)( 022
02
02 y y g s sav vvv y yc
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Projectile Motion
Assumptions:(1) free-fall acceleration(2) neglect air resistance
Choosing the y direction as positive upward:a x = 0; a y = - g ( a constant)
Take x 0= y 0 = 0 at t = 0
Initial velocity v 0 makes anangle 0 with the horizontal
v 0
x
y
v v v v x y0 0 0 0 0 0 cos sin
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Maximum Height
At the peak of its trajectory, v y = 0.From
Time t 1 to reach the peak
Substituting into:
g
vt y01
g
v yh y
2
20
max
00 gt v gt vv oy y y
20 2
1 gt t v y 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 90 o. For maximal distance the optimal angle is 45 o.
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10 degreesProjection angle = 10 degrees
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10 degrees
30 degrees 40 degrees45 degrees
Projection angle = 45 degrees
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10 degrees
30 degrees40 degrees 45 degrees60 degrees
Projection angle = 60 degrees
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10 degrees
30 degrees40 degrees45 degrees 60 degrees 75 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 V 0 = 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 = v 0 cos 0 = (37.0 m/s) cos 53.1 = 22.2 m/s
v0y = v 0 sin 0 = (37.0 m/s) sin 53.1 = 29.6 m/s
a) position
x = v 0xt = (22.2 m/s)(2.00 s) = 44.4 my = v 0yt - gt2
= (29.6 m/s)(2.00 s) (9.80 m/s 2)(2.00 s) 2
= 39.6 m
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Solution (con.)
Velocity vx = v 0x = 22.2 m/s vy = v 0y gt = 29.6 m/s (9.80 m/s 2)(2.00 s) = 10.0 m/s
2.24450.0arctan/2.22 /0.10arctan
/3.24
)/0.10(/2.22 2222
sm sm
sm
sm smvvv y x
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Solution (cont.)
b) Find the time when the ball reaches the highest pointof its flight, and find its height H at this point.
s s m
s m
g
v t
gt v v
y
y y
023809
629
0
2
0
1
10
./.
/.
m
s sm s sm
gt t v H y
7.44
)02.3)(/80.9(21
)02.3)(/6.29(
21
22
210 1
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Solution (cont.)c) Find the horizontal range R, (that is, the horizontaldistance from the starting point to the point at which the
ball hits the ground.)
m s smt v R x 134)04.6)(/2.22(20
)21(
210 2022220 gt vt gt t v y y y
s sm sm
g
vt and t
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 = v 0-gt
y = y 0 + v 0t + 1/2gt 2 .
v2 = v 02 - 2g(y-y 0).
Horizontally
x = x 0 + (v 0)x t x = 25 *3.19 = 79.8 m
79.8 m
Initial Conditions
vx = 25 m/s
vy0 = 0 m/s
a =- 9.8 m/s2
t = 0
y0 = 0 m
y =- 50 m
x0 =0 m
-50 = 0+0+1/2(-9.8)t 2 t = 3.19 s
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Review
Example 12.11 Example 12.12 Example 12.13
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