Chapter 6B – Projectile Chapter 6B – Projectile MotionMotion

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to: Describe the motion of a projectile by Describe the motion of a projectile by

treating horizontal and vertical treating horizontal and vertical components of its position and components of its position and velocity.velocity.

• Solve for position, velocity, or time Solve for position, velocity, or time when given initial velocity and launch when given initial velocity and launch angle.angle.

Projectile MotionProjectile MotionA A projectile projectile is a particle moving near is a particle moving near the Earth’s surface under the the Earth’s surface under the influence of its weight only (directed influence of its weight only (directed downward).downward).

a = gW

WW

Vertical and Horizontal Vertical and Horizontal MotionMotion

Simultaneously Simultaneously dropping dropping yellowyellow ball ball and projecting and projecting red red ball horizontally.ball horizontally.

Click right to observe motion of each ball.

Vertical and Horizontal Vertical and Horizontal MotionMotion

Simultaneously Simultaneously dropping a dropping a yellowyellow ball ball and projecting a and projecting a red red ball ball horizontally.horizontally.

Why do they strike the ground at the same

time?

Once motion has begun, the downward weight is the only force on each ball.

Once motion has begun, the downward weight is the only force on each ball.

W WW W

Ball Projected Horizontally Ball Projected Horizontally and Another Dropped at and Another Dropped at

Same Time:Same Time:

0 svvoxox

Vertical Motion is the Same for Each Vertical Motion is the Same for Each BallBall

1 s

2 s

3 s

vvyy

vvxx

vvxx

vvxx

vvyy

vvyy

vvyy

vvyy

vvyy

Observe Motion of Each Observe Motion of Each BallBall

0 svvoxox

Vertical Motion is the Same for Each Vertical Motion is the Same for Each BallBall

3 s

2 s

1 s

Consider Horizontal and Consider Horizontal and Vertical Motion Vertical Motion

Separately:Separately:Compare Displacements and Compare Displacements and

VelocitiesVelocities0 s0 s

0 s0 s1 s1 svvoxox

2 s2 s 3 s3 s

1 s1 svvyy

2 s2 svvxx

vvyy3 s3 s

vvxx

vvyy

Horizontal velocity doesn’t change.

Vertical velocity just like free fall.

vvxx

Displacement Calculations for Displacement Calculations for Horizontal Projection:Horizontal Projection:

For any constant acceleration:

Horizontal displacement:

oxx v t

Vertical displacement:

212y gt

212ox v t at

For the special case of horizontal projection: 0; 0; x y oy ox oa a g v v v

Velocity Calculations for Velocity Calculations for Horizontal Projection (cont.):Horizontal Projection (cont.):

For any constant acceleration:

Horizontal velocity: x oxv v

Vertical velocity: y ov v gt

f ov v at For the special case of a projectile:

0; 0; x y oy ox oa a g v v v

Example 1:Example 1: A baseball is hit with a A baseball is hit with a horizontal speed of horizontal speed of 25 m/s25 m/s. What . What is its position and velocity after is its position and velocity after 2 2 ss??

First find horizontal and vertical displacements:(25 m/s)(2 s)oxx v t

2 2 21 12 2 ( 9.8 m/s )(2 s)y gt

x = 50.0 m

x = 50.0 m

y = -19.6 m

y = -19.6 m

25 m/s

xy-19.6 m-19.6 m

+50 m+50 m

Example 1 Cont.):Example 1 Cont.): What are the What are the velocity components after velocity components after 2 s2 s??

25 m/s

Find horizontal and vertical velocity after 2 s:

(25 m/s)x oxv v 20 ( 9.8 m/s )(2 s)y oyv v at

vx = 25.0 m/s

vx = 25.0 m/s

vy = -19.6 m/s

vy = -19.6 m/s

vx

vy

v0x = 25 m/s v0y = 0

Consider Projectile at an Consider Projectile at an Angle:Angle:

A red ball is projected at an angle . At the same time, a yellow ball is thrown vertically upward and a green ball rolls horizontally (no friction).

A red ball is projected at an angle . At the same time, a yellow ball is thrown vertically upward and a green ball rolls horizontally (no friction).

Note vertical and horizontal motions of balls

voy

vox

vo

vx = vox = constant

y oyv v at 29.8 m/sa

Displacement Calculations For Displacement Calculations For General Projection:General Projection:

The components of displacement at time t are:

212ox xx v t a t 21

2ox xx v t a t

For projectiles:

0; ; 0; x y oy ox oa a g v v v

212oy yy v t a t 21

2oy yy v t a t

Thus, the displacement components x and y for projectiles are:

212

ox

oy

x v t

y v t gt

Velocity Calculations For Velocity Calculations For General Projection:General Projection:

The components of velocity at time t are:

x ox xv v a t x ox xv v a t

For projectiles:

0; ; 0; x y oy ox oa a g v v v

y oy yv v a t y oy yv v a t

Thus, the velocity components vx and vy for projectiles are:

constantx ox

y oy

v v

v v gt

Problem-Solving Strategy:Problem-Solving Strategy:1.1. Resolve initial velocity vResolve initial velocity voo into into

components:components:vo

vox

voy cos ; sinox o oy ov v v v cos ; sinox o oy ov v v v

2. Find components of final position and 2. Find components of final position and velocity:velocity:

212

ox

oy

x v t

y v t gt

Displacement: Velocity:

0

2

x x

y oy

v v

v v gt

Problem Strategy (Cont.):Problem Strategy (Cont.):

3. The final position and velocity can be 3. The final position and velocity can be found from the components.found from the components.

R

x

y

4. Use correct signs - remember 4. Use correct signs - remember gg is is negative or positive depending on your negative or positive depending on your initial choice.initial choice.

2 2 ; tany

R x yx

2 2 ; tan

yR x y

x

2 2 ; tan yx y

x

vv v v

v

2 2 ; tan yx y

x

vv v v

v

vo

vox

voy

Example 2:Example 2: A ball has an initial A ball has an initial velocity of velocity of 160 ft/s160 ft/s at an angle of at an angle of 3030oo with horizontal. Find its position and with horizontal. Find its position and velocity after velocity after 2 s2 s and after and after 4 s4 s..

voy 160 ft/s

vox30o

Since Since vvxx is constant, the horizontal is constant, the horizontal displacements after 2 and 4 seconds are:displacements after 2 and 4 seconds are:

(139 ft/s)(2 s)oxx v t x = 277 ftx = 277 ft

(139 ft/s)(4 s)oxx v t x = 554 ftx = 554 ft

0(160 ft/s) cos30 139 ft/soxv 0(160 ft/s)sin 30 80.0 ft/soyv

Note:Note: We know ONLY the We know ONLY the horizontal horizontal locationlocation after after 22 and and 4 s4 s. We don’t know . We don’t know whether it is on its way up or on its way whether it is on its way up or on its way down. down.

x2 = 277 ftx2 = 277 ft x4 = 554 ftx4 = 554 ft

Example 2: (Continued)Example 2: (Continued)

voy 160 ft/s

vox30o

277 ft 554 ft

2 s

4 s

Example 2 (Cont.):Example 2 (Cont.): Next we find the Next we find the vertical components of position after vertical components of position after 2 s2 s and after and after 4 s.4 s.

voy= 80 ft/s

160 ft/s

0 s

3 s

2 s

1 s

4 s

g = -32 ft/s2

y2 y4

2 2 21 12 2(80 ft/s) ( 32 ft/s )oyy v t gt t t

The vertical displacement as function of time:

280 16y t t 280 16y t t Observe consistent units.

(Cont.)(Cont.) SignsSigns of of yy will indicate will indicate locationlocation of displacement (above of displacement (above ++ or or below below –– origin). origin).

voy= 80 ft/s

160 ft/s

0 s

3 s

2 s

1 s

4 s

g = -32 ft/s2

y2 y4

Vertical position:

280 16y t t 280 16y t t 2

2 80(2 s) 16(2 s)y 24 80(4 s) 16(4 s)y

2 96 fty 2 96 fty 4 16 fty 4 16 fty

96 ft16 ft

Each above origin Each above origin (+)(+)

(Cont.):(Cont.): Next we find horizontal and Next we find horizontal and vertical components of vertical components of velocityvelocity after after 22 and and 4 s.4 s.

Since vx is constant, vx = 139 ft/s at all times.

Since vx is constant, vx = 139 ft/s at all times. Vertical velocity is same as if vertically

projected: 2; where g 32 ft/sy oyv v gt At any time t:

(32 ft/s)y oyv v t 139 ft/sxv

voy 160 ft/s

vox30o

0(160 ft/s) cos30 139 ft/soxv 0(160 ft/s)sin 30 80.0 ft/soyv

v2y = 16.0 ft/sv2y = 16.0 ft/s

v4y = -48.0 ft/sv4y = -48.0 ft/s

Example 2: (Continued)Example 2: (Continued)vy= 80.0

ft/s 160 ft/s

0 s

3 s

2 s

1 s

4 s

g = -32 ft/s2 v2

v4

At any time t:

(32 ft/s)y oyv v t 139 ft/sxv

80 ft/s (32 ft/s)(2 s)yv

80 ft/s (32 ft/s)(4 s)yv

At 2 s: v2x = 139 ft/s; v2y = + 16.0 ft/s

At 2 s: v2x = 139 ft/s; v2y = + 16.0 ft/s

Example 2: (Continued)Example 2: (Continued)

vy= 80.0 ft/s 160 ft/s

0 s

3 s

2 s

1 s

4 s

g = -32 ft/s2 v2

v4

Moving Up +16

ft/s

Moving down -48

ft/s

The signs of vy indicate whether motion is up (+) or down (-) at any time t.

At 4 s: v4x = 139 ft/s; v4y = - 48.0 ft/sAt 4 s: v4x = 139 ft/s; v4y = - 48.0 ft/s

(Cont.):(Cont.): The displacement The displacement RR22,, is found is found from the from the xx22 and and yy22 component component displacements.displacements.

0 s

2 s

4 s

y2 = 96 ft

x2= 277 ft

R2

2 2R x y tany

x

2 2(277 ft) (96 ft)R 96 fttan

277 ft

R2 = 293 ft

R2 = 293 ft

= 19.10 = 19.10

t = 2 s

(Cont.):(Cont.): Similarly, displacement Similarly, displacement RR44,, is is found from the found from the xx44 and and yy44 component component displacements.displacements.

2 2(554 ft) (64 ft)R 64 fttan

554 ft

R4 = 558 ft

R4 = 558 ft

= 6.590 = 6.590

0 s

4 s

y4 = 64 ft x4= 554

ft

R4

2 2R x y tany

x t = 4

s

(Cont.): (Cont.): Now we find the velocity Now we find the velocity after after 2 s2 s from the components from the components vvx x

and vand vy.y.

2 22 (139 ft/s) (16 ft/s)v

16 fttan

139 ft

v2 = 140 ft/sv2 = 140 ft/s = 6.560 = 6.560

voy= 80.0 ft/s 160 ft/s

0 s

2 s

g = -32 ft/s2 v2

Moving Up +16

ft/s

v2x = 139 ft/sv2y = + 16.0 ft/s

(Cont.) (Cont.) Next, we find the velocity Next, we find the velocity after after 4 s4 s from the components from the components vv4x 4x

and and vv4y.4y.

2 24 (139 ft/s) ( 46 ft/s)v

16 fttan

139 ft

v4 = 146 ft/sv4 = 146 ft/s = 341.70 = 341.70

voy= 80.0 ft/s 160 ft/s

0 s

4 s

g = -32 ft/s2

v4

v4x = 139 ft/sv4y = - 48.0 ft/s

Example 3:Example 3: What are What are maximum heightmaximum height and and range range of a projectile if of a projectile if vvoo = 28 m/s = 28 m/s at 30at 3000??

ymax occurs when 14 – 9.8t = 0 or t = 1.43 s

Maximum y-coordinate occurs when vy = 0:

voy 28 m/s

vox30o

ymaxvy = 0

214 m/s ( 9.8 m/s ) 0y oyv v gt t

vox = 24.2 m/svoy = + 14 m/s

0(28 m/s)cos30 24.2 m/soxv 0(28 m/s)sin 30 14 m/soyv

Example 3(Cont.):Example 3(Cont.): What is What is maximum maximum heightheight of the projectile if v = 28 m/s at of the projectile if v = 28 m/s at 303000??

Maximum y-coordinate occurs when t = 1.43 s:

ymax= 10.0 m

ymax= 10.0 m

voy 28 m/s

vox30o

ymaxvy = 0

vox = 24.2 m/svoy = + 14 m/s

2 21 12 214(1.43) ( 9.8)(1.43)oyy v t gt

20 m 10 my

Example 3(Cont.):Example 3(Cont.): Next, we find the Next, we find the range range of the projectile if v = 28 m/s at of the projectile if v = 28 m/s at 303000..

The range xr is defined as horizontal distance coinciding with the time for vertical return.

voy 28 m/s

vox30o

vox = 24.2 m/svoy = + 14 m/s

Range xr

The time of flight is found by setting y = 0:21

2 0oyy v t gt (continued)

Example 3(Cont.):Example 3(Cont.): First we find the First we find the time of flight time of flight ttrr, then the , then the range range xxrr..

voy 28 m/s

vox30o

vox = 24.2 m/svoy = + 14 m/s

Range xr

12 0;oyv gt

(Divide by t)212 0oyy v t gt

xr = voxt = (24.2 m/s)(2.86 s);

xr = 69.2 m

xr = 69.2 m

2

2(14 m/s);

-(-9.8 m/2.86

s )soy t

vt

g

Example 4:Example 4: A ball rolls off the top A ball rolls off the top of a table of a table 1.2 m1.2 m high and lands on high and lands on the floor at a horizontal distance the floor at a horizontal distance of of 2 m2 m. What was the velocity as it . What was the velocity as it left the table?left the table?

1.2 m2 m

First find t from y equation:

0

½(-9.8)t2 = -(1.2)

t = 0.495 s

t = 0.495 s

Note: x = voxt = 2 m

y = voyt + ½ayt2 = -1.2 m

212 1.2 my gt

2( 1.2)

9.8t

R

Example 4 (Cont.):Example 4 (Cont.): We now use We now use horizontal equation to find horizontal equation to find vvoxox leaving leaving the table top.the table top.

Use t = 0.495 s in x equation:

v = 4.04 m/s

v = 4.04 m/s

1.2 m2 m

RNote: x = voxt = 2 m

y = ½gt2 = -1.2 m

2 moxv t 2 m

(0.495 s) = 2 m; 0.495 sox oxv v

The ball leaves the table with a

speed:

Example 4 (Cont.):Example 4 (Cont.): What will be its What will be its speed when it strikes the floor?speed when it strikes the floor?

vy = 0 + (-9.8 m/s2)(0.495 s)

vy = vy + gt0

vx = vox = 4.04 m/s

Note:t = 0.495

s

vy = -4.85 m/s

2 2(4.04 m/s) ( 4.85 m/s)v 4.85 m

tan4.04 m

v4 = 146 ft/sv4 = 146 ft/s = 309.80 = 309.80

1.2 m2 m vx

vy

Example 5.Example 5. Find the “hang time” for the Find the “hang time” for the football whose initial velocity is 25 m/s, football whose initial velocity is 25 m/s, 606000..

vo =25 m/s

600

y = 0; a = -9.8 m/s2

Time of flight t

vox = vo cos

voy = vo sin

Initial vo:

VVoxox = (25 m/s) cos 60 = (25 m/s) cos 6000; v; voxox = 12.5 = 12.5 m/sm/s

VVoyoy = (25 m/s) sin 60 = (25 m/s) sin 6000; v; voxox = 21.7 = 21.7 m/sm/s

Only vertical parameters affect hang Only vertical parameters affect hang time.time.2 21 1

2 2; 0 (21.7) ( 9.8)oyy v t at t t

vo =25 m/s

600

y = 0; a = -9.8 m/s2

Time of flight t

vox = vo cos

voy = vo sin

Initial vo:

2 21 12 2; 0 (21.7) ( 9.8)oyy v t at t t

4.9 4.9 tt22 = = 21.7 21.7 tt 4.9 4.9 t = t = 21.721.7

2

21.7 m/s

4.9 m/st t = 4.42 st = 4.42 s

Example 5 (Cont.)Example 5 (Cont.) Find the “hang time” for Find the “hang time” for the football whose initial velocity is 25 m/s, the football whose initial velocity is 25 m/s, 606000..

Example 6Example 6. . A running dog leaps with A running dog leaps with initial velocity of 11 m/s at 30initial velocity of 11 m/s at 3000. What is . What is the range?the range?

v = 11 m/s

=300

Draw figure and Draw figure and find find

components:components:vvox ox = = 9.53 m/s9.53 m/s

vvoy oy = = 5.50 m/s5.50 m/s vox = 11 cos 300

voy = 11 sin 300

2 21 12 2; 0 (5.50) ( 9.8)oyy v t at t t

To find range, first find To find range, first find tt when y = 0; when y = 0; aa = -9.8 = -9.8 m/sm/s22

4.9 4.9 tt22 = = 5.50 5.50 tt2

5.50 m/s

4.9 m/st t = 1.12 st = 1.12 s

4.9 4.9 t = t = 5.505.50

Example 6 (ContExample 6 (Cont.) .) A dog leaps with A dog leaps with initial velocity of 11 m/s at 30initial velocity of 11 m/s at 3000. What is . What is the range?the range?

v = 10 m/s

=310

Range is found Range is found from x-from x-

component:component:vvx x = v= voxox = = 9.53 9.53

m/sm/sx = vx = vxxt; t = t; t = 1.121.12

ss

vox = 10 cos 310

voy = 10 sin 310

Horizontal velocity is constant: Horizontal velocity is constant: vvxx = = 9.539.53 m/s m/s

Range: x = 10.7 m

Range: x = 10.7 m

x = (9.53 m/s)(1.12 s) = 10.7 mx = (9.53 m/s)(1.12 s) = 10.7 m

Summary for Projectiles:Summary for Projectiles:

1. Determine x and y components v0

cos and sinox o oy ov v v v cos and sinox o oy ov v v v

2. The horizontal and vertical components of displacement at any time t are given by:

212 ox oyx v t y v t gt 21

2 ox oyx v t y v t gt

Summary (Continued):Summary (Continued):

4. Vector displacement or velocity can then be found from the components if desired:

3. The horizontal and vertical components of velocity at any time t are given by:

; x ox y oyv v v v gt ; x ox y oyv v v v gt

2 2R x y 2 2R x y tan

y

x tan

y

x

CONCLUSION: Chapter 6B CONCLUSION: Chapter 6B Projectile MotionProjectile Motion