LEC 3 Projectile)

8/2/2019 LEC 3 Projectile)

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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



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10 degrees

30 degrees40 degrees45 degrees60 degrees



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10 degrees

30 degrees40 degrees45 degrees60 degrees75 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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