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VERTICAL PROJECTILE MOTION

KEY CONCEPTSThis lesson focuses on the following key concepts:

• Forceofgravity

• Gravitationalacceleration

• Equationsofmotion

TERMINOLOGY & DEFINITIONSAcceleration due to gravity:istheaccelerationexperiencedbyaprojectileduetothegravitational

forcethatactsonit

Gravitational force:istheforceofattractionbetweenanytwoobjectsintheuniverse.Inthecase

ofanobjectonearth,itwillbetheforcewithwhichtheearth,asthebiggerobject,willattractthe

objecttowardsitscentre.Thisgravitationalforceiscalledtheweightofanobjectandismeasured

in Newton (N)

Height:theverticaldisplacementofanobject

X-PLANATION OF KEY CONCEPTS AND TERMINOLOGIES

The Gravitational ForceThegravitationalforceactingontheobjectduringitsmotionisfoundbyusingtheformula:

Fg = ma where: F

gisthegravitationalforceactingontheobject

misthemassoftheobjectand aisaccelerationduetogravity

Thegravitationalforceactingontheprojectileisactuallytheforceofattractionbetweentheearthandtheprojectile.Thustheformulaonthelawofuniversalgravityapplies.Therefore,theforceexperiencedbyaprojectileistheforceofattractionbetweentheearthandtheprojectilewhich

canbefoundusingthefollowingformula:

Fg =

where:Fisthegravitationalforcebetweentheearthandtheprojectile

Misthemassoftheearthand misthemassoftheprojectile

It,therefore,followsthat ma = G .Fromthisequation,itcanbeseenthat

accelerationduetogravity,a = G .Thisequationtellsthattheaccelerationdueto

gravitydoesnotdependonthemassofanobject,butonlyonthemassoftheearthandthedistancebetweentheearthandtheobject.

Mm

r2

Mm

r2

Mm

r2

VerticalProjectileMotion

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Projectile motionVerticalprojectilemotionismotionofanobjectinanupordowndirection.Anobjectinanupordownmotioniscalledaprojectile.Theonlyforcethatactsonaprojectileistheforceduetogravity(gravitationalforce).Thisforceequalstotheweightoftheobject.Sincethemovementofaprojectileisinupordownmotion,itsdisplacementisinaverticaldirectionand,therefore,thesymbolusedforitisΔy.Allprojectilesexperiencethesameaccelerationduringmotion.Thisaccelerationisaccelerationduetogravity,whichis9.8m∙s-2inadownwarddirection(towardsthecentreoftheearth).

Whenanobjectmovesinanupordownmotion,itissaidtobemovinginonedimension.Thefollowingequa-tionsareusedinsolvingverticalprojectilemotionproblems:

Notice that ∆y isusedforthedisplacementsincetheobjectmovesinaverticaldirection.

Graphs of projectile motionTheaccelerationduetogravityforaprojectileisconstantandequaltog=9.8m∙s-2downwards.Thisapplieswhenaprojectileismovingupordown.Wemustassignadirectiontomotion.Itisusualtomakeuppositiveand,therefore,g=-9,8m.s-2Theaccelerationversustimegraphforthismotionisalwaysthesame

asshownbelow:

Astoneisthrownupwardsfromaparticularheighthabovetheground.Asketchgraphofthepositionversus

timeforthemotionofthestoneisdrawnbelow.

Vf = Vi + a∆t V2f = V2

i + 2a∆y

∆y = Vi∆t + a∆t2 12 ∆y = ∆t ( )V

f + V

i

2

VERTICAL PROJECTILE MOTION

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Notice from the graph that the time the stone takes to reach its maximum height above the ground, Δt1, equals Δt2, the time it takes to fall from maximum height and return to the point of projection, h. The graph below is the velocity versus time graph of the stone thrown up from h.

Notice from the graph that when the stone reaches its maximum height, its velocity becomes zero. Also notice that the velocity with which the stone reaches the point of projection equals the velocity with which it was projected. However, the two velocities are in opposing directions. Thus if the velocity of the stone in the upwards direction is taken to be positive, its velocity as it falls back to its point of projection will be negative as shown in the graph.

X-AMPLE 1Anobjectisthrownupwardswithavelocityof24,5m∙s-1andtakes5storeturntothethrower’shand.Usetheinformationtosketchthefollowinggraphs:

a) Displacementvstime

b) Velocityvstime

c) Accelerationvstime

X-AMPLE 2Aboystandingontopofabuildingthrowsaballverticallyupwardsfromaposition,3,5mabovethe

groundwithaninitialvelocityof10m∙s-1.Ignoretheeffectsofairresistanceandanswerthe

followingquestions.

2.1 Whatisthemagnitudeanddirectionoftheaccelerationoftheball.

2.2 Calculatethemaximumheightreachedbytheballabovetheground.

2.3 Whatwasthevelocityoftheballatitsmaximumheight?

2.4 Calculatethetimetakenbytheballtoreachitsmaximumheight.

2.5 Howmuchtimedidtheballtaketoreachitsmaximumpositionandbacktothe

positionfromwhichitwasthrown.

2.6 Calculatethetotaltimetakenbytheballtoreachtheground.

2.7 Calculatethevelocitywithwhichtheballhitstheground.

2.8 Drawaroughsketchofthevelocity-timegraph.Showrelevantpointsonyour

velocityandtimeaxis.

VERTICAL PROJECTILE MOTION

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X-ERCISE1.Aboystandsattheedgeofahighcliff.Hethrowsastoneverticallyupwardswithaninitialvelocityof

9,8m∙s-1.Thestonestrikesthegroundatapointbelowthecliffafter3,5seconds.Thevelocity-timegraph

belowwasobtainedfrommeasurementsmadeduringthemotionofthestone.

GRAPH OF VELOCITY VS TIME

Usetheinformationonthegraphtoanswerthefollowingquestions:

1.1 Calculatetheaccelerationofthestonebetweentimest=2sandt=3s.

1.2 Atwhichtime(s)isthestonemovingataspeedof5m∙s-1?

1.3 Afterhowmanysecondsdoesthestonereachitshighestpoint?

1.4 Determinetheheightoftheclifffromwhichthestonewasthrown.

1.5 Usingthetopofthecliffastheinitialpositionofthestone,sketchtheposition-timegraph

formotionofthestonefromitshighestpointuntilitreachestheground.Onlyindicate

relevanttimevaluesonthex–axis.

VERTICAL PROJECTILE MOTION

Vel

ocit

y (m

.s-1)

Time(s)

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ANSWERS 1.1 10m∙s-2downwards(Note:graphisnotaccurateenoughtoread9,8m.s-2)

1.2 Time:0,5secondsafterbeingthrown;1,5secondsafterbeingthrown

1.3 Afteronesecond

1.4 31,25m

1.5

VERTICAL PROJECTILE MOTION