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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


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10 degrees

30 degrees40 degrees45 degrees 60 degrees 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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lec 04 (projectile 051)

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