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Kinematics in Two Dimensions: Projectile motion Chapter 3

Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

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Page 1: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Kinematics*in*Two*Dimensions:

Projectile*motion

Chapter(3

Page 2: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

Under&the&influence&of&gravity&alone,&an&object&near&the&surface&of&the&Earth&will&accelerate&downwards&at&9.80m/s2.

2sm80.9−=ya 0=xa

constant == oxx vv

Page 3: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

Example:* The$Height$of$a$Kickoff

A$placekicker$kicks$a$football$at$and$angle$of$40.0$degrees$andthe$initial$speed$of$the$ball$is$22$m/s.$$Ignoring$air$resistance,$determine$the$maximum$height$that$the$ball$attains.

Page 4: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

ov

oxv

oyvθ

voy = vo sinθ = 22m s( )sin 40! =14m s

vox = vo sinθ = 22m s( )cos40! =17m scos !

Page 5: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

y ay vy voy t? "9.80'm/s2 0 14'm/s

Page 6: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

y ay vy voy t? "9.80'm/s2 0 14'm/s

vy2 = voy

2 − 2gy y =vy2 − voy

2

−2g

y =0− 14m s( )2

−2 9.8m s2( )= +10 m

Page 7: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

Example:* The$Time$of$Flight$of$a$Kickoff

What$is$the$time$of$flight$between$kickoff$and$landing?

Page 8: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

y ay vy voy t0 "9.80&m/s2 14&m/s ?

Page 9: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

y ay vy voy t0 "9.80&m/s2 14&m/s ?

y = voyt − 12 gt

2

0 = 14m s( ) t − 12 9.80m s2( ) t2

0 = 2 14m s( )− 9.80m s2( ) t

s 9.2=t

Page 10: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

Example:* The$Range$of$a$Kickoff

Calculate$the$range$R$of$the$projectile.

x = voxt= 17m s( ) 2.9 s( ) = +49 m

Page 11: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Example:)Shoot)the)falling)object.A"metal"can"is"dropped"from"a"platform"at"the"same"time"a"gun"locatedat"some"distance"from"the"can"and"at"a"lower"height"tries"to"shoot"it"as"itfalls."If"the"can"is"initially"a"height"h above"the"height"of"the"gun,"the"gunis"a"horizontal"distance"d away"from"the"can,"and"the"muzzle"velocity"of"the"bullet"is"v0,"at"what"angle"! should"the"gun"be"aimed"to"hit"the"falling"can?"

! ?v0

x

y

d

h

Projectile*Motion

Page 12: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

! ?v0

x

y

d

h

If t is the time it takes for the bullet to reach the can ⇒ xBullet = d = v0 xt

∴t = dv0 x

=d

v0 cosθy positions at time t: yBullet = v0 yt − 1

2 gt2 = (v0 sinθ )t − 1

2 gt2 and yCan = h− 1

2 gt2

Condition for the bullet to hit the can at time t⇒ yBullet = yCan(v0 sinθ )t − 1

2 gt2 = h− 1

2 gt2 ⇒ (v0 sinθ )t = h

∴ v0 sinθ( ) dv0 cosθ$

%&

'

()= h ⇒ tanθ = h

d⇒ θ = tan−1 h

d$

%&

'

()

∴ aim the gun at the initial position of the can to hit it while it is falling

Projectile*Motion

Page 13: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

dmin ?

}#hw

vmin ?

q

At#serve,#a#tennis#player#aims#to#hit#the#ball#horizontally.#

a)#What#minimum#speed#is#required#for#the#ball#to#clear#the#h =#0.90#m#high#net#q =#15.0#m#from#the#server#if#the#ball#is#"launched"#from#a#height#of#w =#2.00#m?b)#Where#will#the#ball#land#relative#to#the#player#if#it#just#clears#the#net?c)#How#long#will#the#ball#be#in#the#air?

Projectile*Motion

Example:*Serving#a#tennis#ball.

tmin ?

+x

+y

Page 14: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

dmin ?

}#hw

vmin ?

q

Projectile*Motion

tmin ?

+x

+y

y ay vy voy t$(w/h) $g 0 ?

y = voyt − 12 gt

2 ⇒ t = −2(w− h)−g

=2(1.10)

9.80= 0.474 s

x = q = vmint ⇒ vmin =qt=

15.00.474

= 31.6 m/s

a)

Page 15: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

dmin ?

}#hw

vmin ?

q

Projectile*Motion

tmin ?

+x

+y

y ay vy voy t$w $g 0 ?

y = voyt − 12 gt

2 ⇒ tmin =−2w−g

=2(2)9.80

= 0.639 s

x = dmin = vmintmin = (31.6)(0.639) = 20.2 m

b)#and#c)

Page 16: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*MotionConceptual*Example: Two$Ways$to$Throw$a$Stone

From$the$top$of$a$cliff,$a$person$throws$two$stones.$$The$stoneshave$identical$initial$speeds,$but$stone$1$is$thrown$downwardat$some$angle$below$the$horizontal$and$stone$2$is$thrown$atthe$same$angle$above$the$horizontal.$$Neglecting$air$resistance,which$stone,$if$either,$strikes$the$water$with$greater$velocity?

v2 ?x

y

Page 17: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

Projectile*Motion

Initial'velocity'of'stone'1:'v0 at'angle'3! from'horizontal

Initial'velocity'of'stone'2:'v0 at'angle'+! from'horizontal

Velocity'of'stone'2'at'point'P:''y2 ='0

! Both'stones'strike'the'water'with'the'same'velocity

v0 x1 = v0 cos(−θ ) = v0 cosθ , v0 y1 = v0 sin(−θ ) = −v0 sinθ

v0 x2 = v0 cosθ , v0 y2 = v0 sinθ

vx2 = v0 x2 = v0 cosθ

vy22 = v0 y2

2 − 2gy2 = v0 y22 − 2g ⋅0 = v0 y2

2

vy2 = ±v0 y2 = −v0 y2 = −v0 sinθ

∴v0 x1 = vx2 , v0 y1 = vy2

v2 ?

Page 18: Chapter(3 · Projectile*Motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v osinθ=(22ms)cos40 cos !!=17ms

stone&2&has&the&same&velocity&when&itis&at&P as&stone&1&has&initially,&so

! both&will&hit&the&water&with&the&same&velocity(but&stone&2&will&hit&the&water&later&and&behorizontally&displaced&from&stone&1)