ePhysics 12 for VCE Units 3 & 4_Proj

8/3/2019 ePhysics 12 for VCE Units 3 & 4_Proj

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A projectile is any object that is thrown or projected into the air and ismoving freely, i.e. it has no power source (such as a rocket engine) drivingit. A netball as it is passed, a coin that is tossed and a gymnast performing adismount are all examples of projectiles. If they are not launched verticallyand if air resistance is ignored, projectiles move in parabolic paths.

If a ir res is ta nce is ig no red , th e o nly fo rce ac tin g o n a P R O J E C T I L E d u rin g it s f lig h tis its w eig ht, w hich is th e force d ue to g ra vity , F g o r W .T his fo rc e is co ns ta nt a nda lw a ys d ire cte d v er tic ally d ow n w ar ds , a nd c au se s th e p ro je ctile to c on tin ua llyd ev ia te fro m a stra ig ht lin e p ath to fo llo w a p ara bo lic p ath.

Given that the only force acting on a projectile is the force of gravity, Fg ,it follows that the projectile must have a vertical acceleration of 9.8 m S-2downwards. The only force, Fg , that is acting on a projectile is vertical and so it has noeffect on the horizontal motion. The vertical and horizontal components ofthe motion are independent of each other and must be treated separately.

There are no horizontal forces acting on the projectile, so the horizontalcomponent of velocity will be constant.+ I n th e V E RT IC A L C O M P O NE N T , a p ro je ctile a cc ele ra te s w ith th e a cc ele ra tio n d ue tog ra vity , 9 .8 m s-2downward .I n th e H O R IZ O N T A L C O M PO N EN T , a p ro je ctile m o ve s w ith u nifo rm v elo city s in ceth ere a re n o fo rc es a ctin g in th is d ire ctio n.

These points are fundamental to an understanding of projectile motion,and can be seen by studying Figure 1.20.

If air resistance is ignored, the motion of a projectile will be symmetricalaround the vertical axis through the top of the flight. This symmetry extendsto calculations involving speed, velocity and time as well as position. As seenin Figure 1.21, the projectile will take exactly the same time to reach the topof its flight as it will to travel from the top of its flight to the ground. At anygiven height, the speed of the projectile will be the same, and at any givenheight the velocities are related. On the way up, the angle for the velocityvector will be directed above the horizontal, whereas on the way down,the angle is the same but it is directed below the horizontal. For example, aprojectile launched at 50msat 80above the horizontal will land at 50 msat 80 below the horizontal.

-INm u ltifla sh p ho to o f t w o g olf b alls re le as ed s im u lta ne ou sly . O ne b all w as la un ch edh oriz on ta lly a t 2 .0 m S-1 w h ile th e o th er w a s r ele as ed fr om r es t. T he p ro je ct ile la un ch edh or iz on ta lly t ra ve ls a n e qu al h or iz on ta l d is ta nc e d ur in g e ac h f la sh in te rv al, in dic atin g th at it sh or iz on ta l v elo city is c on st an t. H ow e ve r, in th e v er tic al d ir ec tio n, t his p ro je ctile tr av els g re ate rd is ta nc es a s it fa lls . In o th er w ord s, it h as a v ertic al a cc ele ra tio n. In fa ct, b oth b alls h av e a v ertic ala cc ele ra tio n of 9 .8 m s -', s o th ey b oth fa ll a t exa ctly th e s am e ra te a nd la nd a t t he s am e tim e.T his s ho ws th at th e tw o c om po ne nts o f m o tio n a re in de pe nd en t: th e h oriz on ta l m o tio n o f t hela un ch ed p ro je ctile h as n o e ffe ct o n its v ertic al m o tio n [a nd v ic e v ers a].

m o ve s in a p ar ab olic p ath .

;;~@'l- '-P'-"RA""C'-"TI,-"C""AL"-'A""C'-"TI'- ' -V'- '-IT-'--Y-,


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~ = : : : = = : : : : ~ ~ ~ ~ ~ ~ ~ _ : _ : _ : _ = - : : : : : : : : = = = = = = = =. - - - - - -. . . . . ------------------_ _ _ ~ _~ i i i i I I I I I - _ -_ _-= = - ~ -- ~ - - ~ - - - , - - - ------- - -----------

A common misconception is that thereis a driving force acting to keep aprojectile moving through the air. Thisis a medieval understanding of motion.Such a force does not exist. For example,when you toss a ball across the room,your hand exerts a force on the ball asit is being thrown, but this force stopsacting when the ball leaves your hand.There is no driving force propelling theball along. Only the forces of gravity andair resistance act on the ball once it is inmid-air.

-tng n or in g a ir r es is ta n ce , t he h o ri zo n ta lv elo city o f t he b all w ill r em a in th e s am e , w h ile th ev ertic al c om p on en t o f t he v elo city w ill c ha ng e w ithtim e . T he m o tio n o f th e p ro je ctile is s y mm etr ic a l,a nd fo r a g iv en h eig ht, th e b all w ill h av e th e s am espeed .

It can be shown mathematicallythat the path of a projectile will be aparabola. Consider a projectile launchedhorizontally with speed v, and anacceleration down given by g. Let JCand J II be the horizontal and verticaldisplacements respectively.

At time ~ the horizontal displacementis given by: .x;=vt

The vertical displacement at time tJis:JI I = ~gt2 (2)

Substituting (1) into (2) for time, we get:JI I = ~ g ( ~ )= ( 2 ~ 'X 2

Since 9 and V I do not vary, ;Y DC x', whichis the relationship for a parabola.

Motion in one and two dimensions

(1)

v=50 m 5-1at 80 down

T ips fo r p ro blem s invo lv ing pro jec tile m otion Construct adiagram showing the motion and set the problem out clearly.Distinguish between information supplied for each component of themotion.

In the horizontal component, the velocity, v, of the projectile is constant andso the only formula needed is v = xf t:

For the vertical component, the projectile is moving with a constantacceleration (9.8m S-2 down), and so the equations ofmotion for uniformacceleration must be used.

In the vertical component, it is important to clearly specify whether upor down is the positive or negative direction, and use this consistentlythroughout the problem.

W orke d e xa mp le 1 . S AHorizontal launchAgolf ball of mass 150 g is hi t hor izontal ly f rom the top of a 40.0 m high clif f with a speed of25.0 m s-I.Assuming an acceleration dueto gravity of9.80 m S-2 and ignoring air resistance,calculate:a the t ime that the bal l takes to landb the distance that the ball travels from the base of the cliffC the velocity of the ball as it lands


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---- ----- - .. . -- - .. . -- ------ -- .. . _ - .. . _ - - .. . _ - .. . _ - . . . . . . . . . - .. . _ - .. . - ---- .. . _ _ -- ---- -- .. . _d the net force acting on the ball at points A and Be the acceleration of the ball at points A and B .

B

!g = 9.80 m S-2

a Tof ind the t ime of flight of the ball, you need only consider the vertical component.The instant after it is hit , the ball is travelling only hor izontally, so its ini tial vert icalvelocity is zero. Taking down as the positive direction:Uv = 0, a = 9.80 m s-',x = 40.0 m, t =?Substituting in x =ut + ~at' forthe vertical direction only:40.0 = 0 +0.5 X 9.80 X t', so:

t =) 40.0/(0.5 X 9.80J= 2.86 s

The ball takes 2.86 s to reach the ground.b To f ind the horizontal distance travelled by the ball [i.e. the range of the ball], it is

necessary to use the horizontal component.uh = 25.0 m s', t = 2.86 s.x. =?x h =uhtx h = 25.0 X 2.86

= 71.5 mThe bal l lands 71.5 m from the base of the cli ff [i .e. the range of the ball is 71.5 m].

C Todetermine the velocity of the ball as it lands, the horizontal and vertical componentsmust befound separately and then added asvectors. From(a], the ballhas beenairbornefor 2.86 s when it lands. The horizontal veloci ty of the bal l is constant at 25.0 m S-I.Thevertical component ofvelocity when the ball lands is:Uv= 0, a =9.80 m s-',x = 40.0 m, t = 2.86 s

Substituting in v = u + at forthe vertical direction only:v=0+(9.80 X 2.86J= 28.0 m S-1

The actual velocity, v, of the ball is the vector sum of its vertical and horizontalcomponents, as shown in the diagram. The magnitude of the velocity can be found byusing Pythagoras's theorem:v = )25.0' + 28.0'= - - J 1409= 37.5m S-1

The angle at which it lands can befound by using trigonometry:tanS = 28.0

25.0= 1 .12S =48Z

25.0 m S-1e

28.0 m S-1v

Chapter 1 Motion


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~ = : : : = = : : : : ~ ~ ~ ~ ~ ~ ~ _ : _ : _ : _ = - : : : : : : : : = = = = = = = =. - - - - - -. . . . . ------------------_ _ _ ~ _~ i i i i I I I I I - _ -_ _-= = - ~ -- ~ - - ~ - - - , - - - ------- - -----------

7.50 m S-1

When the ball hits the ground, it has a speed of 37.5 m S-1 and is travelling at an angle of48Z below the horizontal.

d If air resistance is ignored, the only force acting on the ball throughout i ts fl ight is itsweight . Therefore the net force that is acting at point A and point B (and everywhereelsel] is:sr- F g =mg

= 0.150 X 9.80= 1.47 N down

e Since the ball is in free-fall, the acceleration of the ball at all points is equal to thatdetermined by gravity, i .e.9.80 m s-' down.

W orke d e xa mp le 1.58Laun ch at an an gle

A 65 kg athlete in a long-jump event leaps with a velocity of 7.50 m S-1 at 30.00 to thehor izontal. Treating the athlete as a point mass, ignoring air resistance, and using 9 as9.80 m s-', calculate:a the horizontal component of the initial velocityb the vertical component of the initial velocityC the velocity when at the highest pointd the maximum height gained bythe athletee the total t ime for which the athlete is in the airf the horizontal distance travelled by the athlete's centre of mass (assuming that it

returns to its original height]g the athlete's acceleration at the highest point of the jump.

In this problem, the upward direction wil l be taken as positive. The horizontal and verticalcomponents of the initial velocity can be found by using trigonometry.a Asshown in the diagram, the horizontal component, uh ' ofthe athlete's initial velocity is:

uh = 7.50 X cos30.0= 6.50 m s+to the right

This remains constant throughout the jump.b Again referring to the diagram, the vertical component, uv' of the init ial velocity of the

athlete is:Uv = 7.50 X sin30.0

= 3.75 m S-1 upwardsC At the highest point, the athlete is moving horizontally. Thevertical component of the

velocity at this point is therefore zero. The actual velocity is given by the hor izontalcomponent of the velocity throughout the jump. This was found in (a] to be 6.50 m S-1in the horizontal direction.

d Tofind the maximum height that is gained, we must work with the vertical component.As explained in [c J . at the maximum height the athlete is moving horizontally and so



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- - - ~ - ~ - ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~---- - . . . . . . - - .. . -- ------ ---- ------ --- .. . --- . . . . . . - ---- ---- ----- ---- ----the vertical component of velocity at this point is zero. Thevertical displacement of theathlete to the highest point is the maximum height that was reached:v.: 3.75 m s', v= 0, a = -9.80 m s-',x=?v' =u' + 2a xo =3.75'+(2X-9.80 x x]x =0.717 mi.e. the centre of mass of the athlete r ises by a maximum height of 72 cm.e As the motion is symmetrical, the t ime to complete the jump will be double that takento reach the maximum height. First, the time to reach the highest point must be found.Using the vertical component:Uv = 3.75 m s', v= 0, a =-9.80 m s-', t =?v = u + ato =3.75+(-9.80xt]t = 0.383 s

The time for the complete flight is double the time to reach maximum height, i.e. totaltime in the air: l:t = 2 x 0.383 = 0.766 s.

f Tofind the horizontal distanceforthe jump, wemustworkwith the horizontal component.From part e, the athlete was in the air for a t ime of 0.766 s and so:t =0.766s,v=6.50ms-\x=?xv=T'sOx=v X t= 6.50 X 0.766=4.98 mi.e.the athlete jumps a horizontal distance of 4.98 m.g At the highest point of the motion, the only force acting on the athlete is that due togravity [i.e. weight]. Theacceleration wil l therefore be 9.80 m s-, down.

The e ffec t o f air res is tanceInthrowing events such as the javelin and discus, new records are not acceptedifthe wind isproviding too much assistance to the projectile. In football games,kicking with the wind is generally an advantage to a team; and in cricket,bowling with the wind, across the wind or against the wind can have verydifferent effects on the flight ofthe ball. The interaction between aprojectile andthe air can have a significant effect on the motion of the projectile, particularlyif the projectile has a large surface area and a relatively low mass.[a ] [b ]

~-~O.FgF ~~

9 Fg '-E _ )..} ~~~l'(r . ' i= , ~.~ Fg " " o r air resistance

Eg

~\

with air resistanci'\{.al\ ~) l

A conservation of energy approachcan also be used for solving projectilesproblems. When air resistance can beignored, the sum of the gravitationalpotential energy and kinetic energyof the projectile (i.e, its mechanicalenergy) is the same at all points inits flight. At the lowest point in itsflight, gravitational potential energyis a minimum and kinetic energy is amaximum. At the highest point, theopposite occurs. The mass needs to begiven to determine the actual energyvalues, but is not required to find theother properties such as acceleration,speed and displacement. Energy will bediscussed in detail in Chapter 2.

!mtf' raj Thepathof a food parceldroppedfrom a plane.Ifthe planemaintains a constant speedandin the absenceof air resistance,the parcelwillfall in a parabolicpath andremain directly belowthe plane.Airresistancemakesthe parcelfall moreslowly, overa shorter path. [b] Whenair resistance isacting, the net forceon the parcelis notvertically down.

Chapter 1 Motion


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~ = : : : = = : : : : ~ ~ ~ ~ ~ ~ ~ _ : _ : _ : _ = - : : : : : : : : = = = = = = = =. - - - - - -. . . . . ------------------_ _ _ ~ _~ i i i i I I I I I - _ -_ _-= = - ~ -- ~ - - ~ - - - , - - - ------- - -----------Figure 1.23 shows a food parcel being dropped from a plane moving at

a constant velocity. If air resistance is ignored, the parcel falls in a parabolicarc. It would continue moving horizontally atthe same rate as the plane; thatis, as the parcel falls it would stay directly beneath the plane until it hits theground. The effect of air resistance is also shown. Air resistance (or drag) isa retarding force and it acts in a direction that is opposite to the motion ofthe projectile. Ifair resistance is taken into account, there are now two forcesacting-weight, F g , and air resistance, F a ' Therefore, the resultant force, lJF,that acts on the projectile is not vertically down. The magnitude of the airresistance force is greater when the speed of the body is greater.

The grande jete is a baJlet movement inwhich dancers leapacross the stage and appear to float inthe air for a period oftime. They position their arms and legs to give the impressionthat they are floating gracefully through the air. The dancer'scentre of mass follows a parabolic path. Once the dancer is inmid-air, there is nothing that he or she can do to alter this path.

(b J

Projectiles move in parabolic paths that can beanalysed by considering the horizontal and verticalcomponents of the motion.

If air resistance is ignored, the only force acting ona projectile is its weight, i.e. the force of gravity, Fgor W. This results in the projectile having a verticalacceleration of 9.8 m S-2 down during its flight.

The horizontal speed of a projectile remains constantthroughout its flight if air resistance is ignored.


However, by raising their arms and legs, dancers can raisethe position of their centre of mass so that it is higher in thetorso. The effect of this is that the dancer's head follows alower and flatter line than it would have taken if the limbshad not been raised during the leap. The smoother and flatterline taken by the head gives the audience the impression of agraceful floating movement across the stage.

"path of centreof mass (parabolic)

An object initially moving horizontally, but free to fall,will fall at exactly the same rate, and in the same time,as an object falling vertically from the same height.

At the point of maximum height, a projectile ismoving horizontally. Its velocity at this point is givenby the horizontal component of its velocity as thevertical component equals zero.

When air resistance is significant, the net force actingon a projectile will not be vertically down, nor will itsacceleration. Under these conditions, the path of theprojectile is not parabolic.


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- - - ~ - ~ - ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~---- - . . . . . . - - .. . -- ------ ---- ------ --- .. . --- . . . . . . - ---- ---- ----- ---- ----

For the following questions, assume that the accelerationdue to gravity is 9.8m S-2 and ignore the effects of airresistance unless otherwise stated.1A golfer practising on a range with an elevated tee4.9 m above the fairway is able to strike a ball so that itleaves the club with a horizontal velocity of 20m s'.a How long after the ball leaves the club will it land

on the fairway?b What horizontal distance will the ball travel before

striking the fairway?c What is the acceleration of the ball 0.50s after being

hit?d Calculate the speed of the ball 0.80 s after it leaves

the club.e With what speed will the ball strike the ground?

fairway2 Abowling ball ofmass 7.5kg travelling at 10m S-l rollsoff a horizontal table 1.0m high.a Calculate the ball's horizontal velocity just as it

strikes the floor.b What is the vertical velocity of the ball as it strikes

the floor?c Calculate the velocityof the ball as it reachesthe floor.d What time interval has elapsed between the ball

leaving the table and striking the floor?e Calculate the horizontal distance travelled by the

ball as it falls.f Draw a diagram showing the forces acting on the

ball as it falls towards the floor.The following information applies to questions 3-8.A senior physics class conducting a research project onprojectile motion constructs a device that can launcha cricket ball. The launching device is designed so thatthe ball can be launched at ground level with an initialvelocity of 28msat an angle of 30 to the horizontal.

ground level

3 Calculate the horizontal component of the velocity ofthe ball:a initiallyb after 1.0sc after 2.0s.

4 Calculatetheverticalcomponent ofthe velocityof theball:a initiallyb after 1.0sc after 2.0s.

5 a At what timewilltheball reachitsmaximum height?b What is the maximum height that is achieved by the

ball?c What is the acceleration of the ball at its maximum

height?6 a At which point in its flight will the ball experience

its minimum speed?b What is the minimum speed of the ball during its

flight?c At what time does this minimum speed occur?d Draw a diagram showing the forces acting on the

ball at the maximum height.7 a At what time after being launched will the ball

return to the ground?b What isthevelocityof the ball as itstrikes the ground?c Calculate the horizontal range of the ball.

8 If the effects of air resistance were taken into account,which one ofthe following statements would be correct?A The ball would have travelled a greater horizontal

distance before striking the ground.B The ball would have reached a greater maximum

height.C The ball's horizontal velocity would have been con-tinually decreasing.

9 A softball of mass 250 g is thrown with an initialvelocity of 16m S-l at an angle 9 to the horizontal.When the ball reaches its maximum height, its kineticenergy is 16 J .a What is the maximum height achieved by the ball

from its point of release?b Calculate the initial vertical velocity of the ball.c What is the value of 9?d What is the speed of the ball after 1.0s?e What is the displacement of the ball after 1.0 s?f How long after the ball is thrown will it return to

the ground?g Calculate the horizontal distance that the ball will

travel during its flight.10 During training, an aerial skier takes off from a ramp

that is inclined at 40.0 to the horizontal and lands ina pool that is 10.0m below the end of the ramp. If shetakes 1.50s to reach the highest point of her trajectory,calculate:a the speed at which she leaves the rampb the maximum height above the end of the ramp

that she reachesc the time for which she is in mid-air.

Chapter 1 Motion


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~ : : : : : : : : : : ~ ~ ~ ~ ~ ~ : : ~ _ : _ : _ : : _ : - : : : : : : : : : : = = = = = = ~. - - - - -. . . . . ------------------_ _ _ ~ _~ i i i i I I I I I - _ -_ _-= = - ~ -- ~ - - ~ - - - , - - - ------- - -----------For the following quest ions, assume that the acceleration due togravity is 9.8 m S-2 and ignore the effects of air resistance unlessotherwise stated.The following information applies to questions 1 and 2.David is jumping on his trampol ine. His mass is 25 kg and he landsvertically at 3.5 m S-1 before rebounding vertically at 3.0 m S-I.1 a What is David's change of speed as he bounces?b What is David's change of velocity as he bounces?

2 Eachtime David bounces, the trampoline exerts a force on him.a Discuss this force and how itvaries during the short duration

of each bounce. What name is usually given to this force?b Eacht ime that David bounces, he exerts a downwards forceon the trampoline. Which one of the following statements is

correct?AI The force that David exerts on the trampoline is always

greater than the force that the trampoline exerts onDavid.

a The force that David exerts on the trampoline is alwaysless than the force that the trampoline exerts on David.

C Theforce that David exerts on the trampoline is at timesgreater than and at other t imes less than the force thatthe trampoline exerts on David.

D The force that David exerts on the trampoline is alwaysequal to the force that the trampoline exerts on David.

3 An Olympic archery competi tor tests a bow by fir ing an arrow ofmass 25 g vertically into the air .The arrow leaves the bow withan initial vertical velocity of 100 m S-I.a At what time will the arrow reach its maximum height?b What is the maximum vertical distance that this arrow

reaches?c What is the acceleration of the arrow when it reaches its

maximum height?4 A motorcyclist is travelling at 60 km h when she has to braketo a sudden stop. Sheskids and stops in a distance of 15 m. Thecombined mass of the bike and r ider is 120 kg.a What was the magnitude of her average acceleration as she

stopped?b How fast was she travelling after skidding for 1.5 s?c Determine the magnitude of the average retarding force

(in kN J acting on her motorbike as it stopped.5 An aeroplane is headed due north at 500 km h' in sti ll a ir.Thenthe wind starts to blow at a speed of 100 km h towards thewest.

a What is the speed of the plane relative to the ground now?A 600 km ha 400 km h'C 510 kmh'D 490 km h

b Thep ilot wishes to travel due north at 500 km h, In whichdirection and with what air speed should the pilot fly theplane to achieve this goal?A 500 km h' northa 510 kmh' northC 490 km h at 110trueD 510 km h at 110 true

6 Twoidentical tennis balls Xand Yare hit horizontally from a point2.0 m above the ground with different initial speeds: ball X hasan initial speed of 5.0 m S-1 whi le bal l Y has an init ial speed of7. 5 m S-I.a Calculate the time it takes for each bal l to str ike the ground.b Calculate the speed of ball X just before it strikes the

ground.c What is the speed of ball Y just before it strikes the ground?d How much further than ball X does ball Y travel in the

horizontal direction before bouncing?The following information applies to questions 7-10.Thediagram shows the trajectory of aVortex after it has been thrownwith an initial speed of 10.0 m S-I. TheVortex reaches its maximumheight at point 0, 4.00 m higher than its starting height.

V= 10.0 m S-1

7 What is the value of the angle e that the initial velocity vectormakes with the horizontal?8 What is the speed of the Vortex at point O?9 What is the accelerat ion of the Vortex at point O?

A zeroa 9.8 m S-2 forwardsC 4.9 m S-2 downD 9.8 m S-2 down

10 How far away isthe Vortex when it reaches point R?



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- - - ~ - ~ - ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~---- - . . . . . . - - .. . -- ------ ---- ------ --- .. . --- . . . . . . - ---- ---- ----- ---- ----The following information applies to questions 11-15.Ina shot-put event a 2.0 kg shot is launched from a height of 1.5 m,with an initial velocity of 8.0 m S-1 at an angle of 600 to thehorizontal.11 a What is the initial horizontal speed of the shot?

b What is the initial vertical speed of the shot?c How long does it take the shot-put to reach its maximum

height?d What is the maximum height from the ground that is reached

by the shot?e How long after being thrown does the shot reach the

ground?Calculate the total horizontal distance that the shot travelsduring its flight.

12 What is the speed of the shot when it reaches its maximumheight?

13 What is the minimum kinetic energy of the shot during itsflight?

14 What is the acceleration of the shot at its maximum height?15 Which of the following angles of launch will result in the shot

travelling the greatest horizontal distance before returning to itsinitial height?A 150B 300C 450o 600

The following information applies to questions 16-18.During a volleyball match, Adam served the ball of mass 140g froma height of 2.0 m. The ball was served at 12m S-1 and reached amaximum height of 4.5 m above the court.

14.5 mj

16 What was the initial kinetic energy of the volleyball?A 10JB 20 JC 10000 Jo 2.7 J

17 What was the initial gravitational potential energy of thevolleyball?

18 How fast was the ball travelling at its highest point?A zeroB 9.6 m S-1C 9.8 m S-1o 92 m S-1

The following information applies to questions 19and 20.A student at the Australian Inst itute of Sport was able to establishthat during its flight, a 2.0 kg shot experienced a force due to airresistance that was proportional to the square of its speed. Theformula F~= 3.78 X 1O-5v' was determined, where F~is the force dueto air resistance and V 1 is the instantaneous speed of the shot. Theshot-put in one particulartoss was launched at 7.5m S-1 at an angleof 360 to the horizontal.19 Calculate the maximum force due to air resistance that the shot

experiences during its flight.20 Calculate the value ofthe ratio ofthe forces acting onthe shot as

it is tossed:F gF~(max]

What does your answer tel l you about these forces?

Chapter 1 Motion

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ePhysics 12 for VCE Units 3 & 4_Proj