Kinematics*in*Two*Dimensions: Projectile*motion

Transcription

1 Chapter(3 Kinematics*in*Two*Dimensions: Projectile*motion

2 Under&the&influence&of&gravity&alone,&an&object&near&the& surface&of&the&earth&will&accelerate&downwards&at&9.80m/s. a = 9.80m s = 0 y a x v = v x ox = constant

3 Example:* A"Falling"Care"Package The"airplane"is"moving"horizontally"with"a"constant"velocity"of" +115"m/s"at"an"altitude"of"1050"m.""Determine"the"time"required for"the"care"package"to"hit"the"ground.

4 y a y v y v oy t!1050%m!9.80%m/s 0%m/s?

5 y a y v y v oy t!1050%m!9.80%m/s 0%m/s? y = v t + oy 1 a t y y = 1 a t y t = y a y = ( 1050 m) 9.80m s = 14.6 s

6 Example:* The$Velocity$of$the$Care$Package What$are$the$magnitude$and$direction$of$the$final$velocity$of the$care$package?

7 y a y v y v oy t!1050%m!9.80%m/s? 0%m/s 14.6%s

8 y a y v y v oy t!1050%m!9.80%m/s? 0%m/s 14.6%s ( ) 14.6 s v y = v oy + a y t = m s = 143m s ( )

9 Find%the%final%velocity%of%the%package. v = v x ˆx + v y ŷ = (115 m/s)ˆx + ( 143 m/s)ŷ =%5 143%m/s v = v x + v y = (115 m/s) + (-143 m/s) = (m/s)! v = 184 m/s! = tan -1 (-143/115) = -51. o

10 Example:* The$Height$of$a$Kickoff A$placekicker$kicks$a$football$at$and$angle$of$40.0$degrees$and the$initial$speed$of$the$ball$is$$m/s.$$ignoring$air$resistance,$ determine$the$maximum$height$that$the$ball$attains.

11 v o v oy θ v ox v = v sinθ = ( m s)sin 40! =14m s oy o v = v sinθ = ( m s)cos40! =17m s ox o cos!

12 y a y v y v oy t? "9.80'm/s 0 14'm/s

13 y a y v y v oy t? "9.80'm/s 0 14'm/s v = v gy y oy y = v v y oy g y = 0 14m s ( ) 9.8m s ( ) = +10 m

14 Example:* The$Time$of$Flight$of$a$Kickoff What$is$the$time$of$flight$between$kickoff$and$landing?

15 y a y v y v oy t 0 "9.80&m/s 14&m/s?

16 y a y v y v oy t 0 "9.80&m/s 14&m/s? y = v oy t 1 gt 0 = ( 14m s)t 1 ( 9.80m s)t 0 = ( 14m s) ( 9.80m s )t t =.9 s

17 Example:* The$Range$of$a$Kickoff Calculate$the$range$R$of$the$projectile. x = v ox t = ( 17m s) (.9 s) = +49 m

18 Example:)Shoot)the)falling)object. A"metal"can"is"dropped"from"a"platform"at"the"same"time"a"gun"located at"some"distance"from"the"can"and"at"a"lower"height"tries"to"shoot"it"as"it falls."if"the"can"is"initially"a"height"h above"the"height"of"the"gun,"the"gun is"a"horizontal"distance"d away"from"the"can,"and"the"muzzle"velocity"of"the" bullet"is"v 0,"at"what"angle"! should"the"gun"be"aimed"to"hit"the"falling"can?" y h v 0!? x d

19 If t is the time it takes for the bullet to reach the can x Bullet = d = v 0 x t t = d v 0 x = d v 0 cosθ y positions at time t: y Bullet = v 0 y t 1 gt = (v 0 sinθ)t 1 gt and y Can = h 1 gt Condition for the bullet to hit the can at time t (v 0 sinθ)t 1 gt = h 1 gt (v 0 sinθ)t = h $ d ' ( v 0 sinθ) & ) = h tanθ = h % v 0 cosθ ( d aim the gun at the initial position of the can to hit it while it is falling y y Bullet = y Can $ θ = tan 1 h ' & ) % d ( h v 0!? x d

20 w +y +x v min? Example:*Serving#a#tennis#ball. q d min? }#h t min? 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 =#.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?

21 +y +x v min? w q }#h t min? d min? a) y a y v y v oy t $(w/h) $g 0? y = v oy t 1 gt t = (w h) g x = q = v min t v min = q t = = (1.10) 9.80 = s = 31.6 m/s

22 +y +x v min? w q }#h t min? d min? b)#and#c) y a y v y v oy t $w $g 0? y = v oy t 1 gt t min = w g = () 9.80 = s x = d min = v min t min = (31.6)(0.639) = 0. m

23 Conceptual*Example: Two$Ways$to$Throw$a$Stone From$the$top$of$a$cliff,$a$person$throws$two$stones.$$The$stones have$identical$initial$speeds,$but$stone$1$is$thrown$downward at$some$angle$below$the$horizontal$and$stone$$is$thrown$at the$same$angle$above$the$horizontal.$$neglecting$air$resistance, which$stone,$if$either,$strikes$the$water$with$greater$velocity? y x v?

24 Initial'velocity'of'stone'1:'v 0 at'angle'3! from'horizontal v 0 x1 = v 0 cos( θ) = v 0 cosθ, v 0 y1 = v 0 sin( θ) = v 0 sinθ Initial'velocity'of'stone':'v 0 at'angle'+! from'horizontal v 0 x = v 0 cosθ, v 0 y = v 0 sinθ Velocity'of'stone''at'point'P:''y ='0 v x = v 0 x = v 0 cosθ v y = v 0 y gy = v 0 y v y = ±v 0 y = v 0 y = v 0 sinθ g 0 = v 0 y v? v 0 x1 = v x, v 0 y1 = v y! Both'stones'strike'the'water'with'the'same'velocity

25 stone&&has&the&same&velocity&when&it is&at&p as&stone&1&has&initially,&so! both&will&hit&the&water&with&the&same&velocity (but&stone&&will&hit&the&water&later&and&be horizontally&displaced&from&stone&1)