chapter(3 · projectile*motion v o v ox v oy θ v oy =v o sinθ=(22ms)sin40!=14ms v ox =v...
TRANSCRIPT
Kinematics*in*Two*Dimensions:
Projectile*motion
Chapter(3
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
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.
Projectile*Motion
ov
oxv
oyvθ
voy = vo sinθ = 22m s( )sin 40! =14m s
vox = vo sinθ = 22m s( )cos40! =17m scos !
Projectile*Motion
y ay vy voy t? "9.80'm/s2 0 14'm/s
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
Projectile*Motion
Example:* The$Time$of$Flight$of$a$Kickoff
What$is$the$time$of$flight$between$kickoff$and$landing?
Projectile*Motion
y ay vy voy t0 "9.80&m/s2 14&m/s ?
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
Projectile*Motion
Example:* The$Range$of$a$Kickoff
Calculate$the$range$R$of$the$projectile.
x = voxt= 17m s( ) 2.9 s( ) = +49 m
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
! ?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
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
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)
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)
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
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 ?
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)