5 projectile motion 2014

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Physics Unit 3 2014 5. Projectile Motion 1 of 14

5. Projectile Motion

Study Design

investigate and analyse the motion of projectiles near Earths surface including a qualitativedescription of the effects of air resistance;

Introduction

.A projectile is a body that has been thrown or projected, and is travelling freely through the air.While undergoing projectile motion the object is under the constant unbalanced force of gravity.When we study projectile motion, no consideration is given to the force projecting the body or towhat happens when it lands. Air resistance is considered to be negligible (in quantitative

questions), but can be taken into consideration in qualitative questions. The projectile will traveleither: vertically or inclined, depending on the initial angle of projection.

The vertical and horizontal motions are treated independently. Usually the vertical motion is treatedfirst, since it determines how long the projectile is in the air.

Vertically Inclined

For both the vertical and inclined projectiles:

the only force acting is the weight, ie. the bodies are in free fall

acceleration is always10 ms-2downward, (including the point C)

the instantaneous velocity is tangential to the path the total energy (KE & PE) is constant

between any two points KE = - PE

paths are symmetrical for time eg. t (A to B) = t (D to E):

t (A to C) = t (C to E)

paths are symmetrical for speed eg. speed at A = speed at E; speed at B = speed at D.

for vertical motion vc= 0, for inclined motion vc 0.

Inclined or oblique projections

the only force acting is vertically down, so the acceleration and change in velocity are vertical.

horizontally there is no component of force, so constant horizontal velocity.

Maximum range is when angle of projection is 450

Ground Level Ground Level

A

B

A

C C

BD D

E E


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Horizontal projection

When the body is launched horizontally and follows a parabolic path to the ground it is really thesecond half of an inclined projection. The time of flight is controlled by the height from which it isreleased. The speed of projection will not affect the time 't' that it takes to land. The 'range' of thisprojectile is given by the x = vhorizontal t.

For projectiles thrown horizontally and dropped from rest, the vertical motions are the same. Thiscan be shown by a multi-flash photograph.

The interval between the lines represents how farthe ball has travelled in a small time interval.Notice that the distance between the horizontallines increases as the ball descends, indicatingthat the ball is speeding up. The distancebetween successive vertical lines remainsconstant, indicating that the ball is travelling witha constant velocity in the horizontal direction.

If we analyse the motion by using resolution of vectors we get the following:

Horizontal: Vertical:velocity always = vhorizontal Velocity v= u + gtacceleration = 0 acceleration = gdisplacement = x = vhorizontal t displacement y = ut + gt

2

To find the 'total' velocity, add vverticaland vhorizontalusing vectors.

If one projectile was fired horizontally, at the same time that another was dropped (from the sameheight), then both objects would hit the ground at the same time. This is because both their verticalmotions were identical. (Same distance to fall, initial speed = 0, and acceleration = -g)

x

y

Uhorizontal

VhorizontalUvertical

Vvertical


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Inclined Projectiles.

ground

The vector representing the initial velocity can be resolved into two componentsv0

v0sin

v0cos

Horizontal: velocity always = vhorizontalv

horizontal= v

0cos

acceleration = 0displacement = x = v0cos t

Vertical(on the way up) velocity changing v = u - gtVvertical= v0sin - gt

acceleration = -g

Vertical(on the way down) velocity changing v= u + gtv = 0 + gt

acceleration = g

The displacement at any time of the motion is: y = ut - gt2

(The same on the way up and down).

Graphs for projectile motion

motion in the 'x' direction motion in the 'y' direction(right is positive) (up is positive)

vo

acceleration acceleration

velocity velocity

displacement displacement


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Symmetrical flightsIf there is no air resistance, and the projectile starts and ends at the same height, then the range is

given by: R =2v sin2

g R is the range, v is the initial speed and the angle of

projection. Be careful using this formula, because it only works under the conditions specified

above.

Total Energy (TE)

If air resistance is negligible, then the total energy of a projectile will always remain

constant throughout the flight. TE = KE + PE = 21

2mv + mgh .

At ground level PE = 0, so TE = KE.

As the projectile rises it gains PE, so it must lose KE. At the top of its flight, the PE ismaximum and the KE is minimum. (the KE is not zero, because the projectile still has someKE due to its horizontal motion).

At any point on the way up or the way down, the TE is constant.

If you know the horizontal component of the velocity, then you can use this to find themaximum height.

Use the TE at ground level and work out what the PE must be at the top when vvertical = 0,but vhorizontal = constant.

Projectile problem methodology1. Draw a fully labelled diagram

2. Treat the motion as two separate motions, for horizontal v0cos =

for vertical, acceleration = -g, v = v0sin

3. List the data under vertical and horizontal

4. For vertical motion use, v2

=u2

+ 2ax, and v = u + at. (use a = -g)5. Usually given information about one direction and asked to find out something about the other6. To go from one direction to another, the common link is the time of flight, t.7. Remember that it will take the same time to go up as to come down.8. Label the direction (+ or -) for all variables except time.


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Su and Harry go for a days surfing with surfboards strapped to the roof of their car. They enter thecliff-top car-park and brake suddenly as they park their car. A surfboard breaks free from the roofstrap, and continues forward in a horizontal direction with an initial speed of 10 m s -1.When it breaks free the surfboard is initially 80 m above the surface of the sea.

Example 1 2000 Question 4Use one or more of Newtons laws to explain why the surfboard continues forward after the carbrakes suddenly.

Example 2 2000 Question 5The path of the surfboard as it continues over the cliff may be modelled as projectile motion withoutair resistance. Calculate the time from the instant the surfboard breaks free until it hits the water.

Example 3 2000 Question 6Calculate the horizontal distance travelled by the surfboard, from where it breaks free to where itfirst strikes the water.


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A car takes off from a ramp and the path of its centre of mass through the air is shown below.

First, model the motion of the car assuming that air resistance is small enough to neglect.

Example 4 1999 Question 8Which one of the directions (AH) best shows the direction of the velocity of the car at point X?

Example 5 1999 Question 9Which one of the directions (AH) best shows the direction of the velocity of the car at point Y?

Example 6 1999 Question 10Which one of the directions (AH) best shows the direction of the acceleration of the car at pointX?

Example 7 1999 Question 11Which one of the directions (AH) best shows the direction of the acceleration of the car at pointY?

Now, suppose that air resistance cannot be neglected.Example 8 1999 Question 12Which one of the directions (AH) could be the direction of the acceleration of the car at point X?


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A skateboarder rides up a ramp as shown below. At the instant the skateboard leaves the ramp thecentre of mass of the skateboard and rider is 1.20 m above the ground and is initially moving with aspeed of 5.0 ms-1at an angle of 30 above the horizontal.The parabolaXYZ is the path of the centre of mass of skateboard and rider.In the following calculations assume that air resistance is negligible.

When the skateboarder is at the highest point of the motion (Y), the speed of the centre of mass is4.33 ms-1.

Example 9 1998 Question 11Calculate the height above the ground of the centre of mass at the highest point of the motion (Y).

When the skateboard touches the ground the centre of mass has moved a horizontal distance of2.87 m from pointX.

Example 10 1998 Question 12Calculate the total time for the centre of mass to travel fromX to Z. (Give your answer to threesignificant figures.)


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Two projectiles Pand Q, each of mass 2.0 kg are given initial horizontal velocities from a point1.8 m above the floor. The path of each projectile is shown in the diagram.Assume air resistance is zero and take g= 10 ms-2.

Projectile initial horizontal velocityP 5.0 ms-1Q 2.0 ms-1

Example 13 1985 Question 5Calculate the KE, in Joules, of Pimmediately before it strikes the floor.

Example 14 1985 Question 6Calculate the value of the ratio

time of flight of

time of flight of

P

Q

Example 15 1985 Question 7Calculate the horizontal distance travelled by Pbefore it strikes the floor.


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A marble of mass Mis projected at an angle from the top of a cliff of height H. At the instant ofprojection it has a kinetic energy of E. This situation is shown below together with the path of theprojectile. Neglect air resistance.

Example 16 1984 Question 6

What is the kinetic energy of the marble at point P?

Example 17 1984 Question 7Write an expression in terms of E, M, gand Hfor the speed of the marble at the pointX.

A ball is shot off a ledge horizontally with a speed of 6.0 ms-1.The ledge is 5.0 m above the ground. (Take g= 10 ms-2)

Example 18 1982 Question 5How long after the ball leaves the ledge does it strike the ground?

Example 19 1982 Question 6How far horizontally from the ledge will the ball strike the ground?


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A boy throws a ball of mass 0.20 kg with an initial speed of 12 ms-1from a height of 1.4 m abovethe ground. It travels in a path as shown in the diagram, reaching a maximum vertical height of3.0m above the starting point.X is the highest point the ball reaches.Take g= 10 Nkg-1and neglect air resistance.

Example 20 1980 Question 4What is the kinetic energy of the ball immediately after it leaves the boys hand?

Example 21 1980 Question 5By how much does the potential energy at pointXexceed the potential energy of the ball as itleaves the boys hand?

Example 22 1980 Question 6What is the kinetic energy of the ball at pointX?

Example 23 1980 Question 7What is the speed of the ball at pointX?

Example 24 1980 Question 8What is the direction of the acceleration of the ball at the point X?A horizontal, to the rightB horizontal, to the leftC vertically upD vertically downE in the directionXto YF no direction, as the acceleration is zero


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SolutionsExample 1 2000 Solution Q4Newtons first law will explain the motion, andthat is a body in motion will continue inmotion unless a net force acts on it. Thesurfboards were travelling at 10ms-1and were

not strapped down therefore they continuedin motion, as there was no net force in thehorizontal plane acting on it.

Example 2 2000 Solution Q5Since the time taken in this example isdetermined by the vertical plane, write downall the values you know for the vertical plane.x = 80mu = 0 ms-1v = ?a = 10 ms-2

t = ?Then you find one of the equations thatinvolves the three known values and will giveyou the unknown value. In this case:

21x = ut + at2

2180 = 0 t + 10 t2

280

t =10

t = 4.0 s (ANS)

Example 3 2000 Solution Q6Since you now have the time take to hit theground, and you know that it was travelling at

10ms-1, you simply use:x

v =t

Manipulate to get:x = v t

and sub the values in.x = 104.04 x = 40.4m (ANS)

Example 41999 Solution Q8C (ANS)

Example 5 1999 Solution Q9D (ANS)

Example 6 1999 Solution Q10The only force acting is the weight force, sothe acceleration is down.

E (ANS)

Example 7 1999 Solution Q11E (ANS)

Same as previous question

Example 8 1999 Solution Q12

F (ANS)Now there are two forces acting, the weightforce down and the air resistance that isopposing the motion.

Example 91998 Solution Q11In the vertical direction the initial velocity isvsin300= 5 0.5 = 2.5 m/s.Use v2= u2- 2gh, to find the height.Since v = 0, (in the vertical direction)

0 = 2.52- 2 10 h

h = 0.3125 mYou must add this to the height that it started

from. h = 1.2 + 0.32= 1.52m (ANS)

Example 101998 Solution Q12Since at the top of the flight, the speed is

4.33 m/s. Then v =d

t

t =d

v

=

2.87

4.33

= 0.663 sec (ANS)

Note that the question specifies the numberof sig.fig. that you must include in youranswer. The zero at the start of the answer isnot significant.

Example 11 1997 Solution Q11Note that the examiners use g = 10N/kg.

Consider the vertical component of thevelocity.

v = u - gtwhere v = 0 and u = 22 sin110

0 = 22sin110- 10 t

t = 4.198 10

t = 0.42

0.42 sec (ANS)


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Example 12 1997 Solution Q12YesYou need to assume that the other part of thebridge is the same height as the first side.If it takes 0.428 sec to get to the top, then itwill take 0.8566 sec to be back to the original

height. The horizontal distance travelled inthis time

= vcos11 0.8566= 22cos 11 0.8566= 18.50 m (ANS)

Example 131985 Solution Q5The total energy of Premains constant.

PE + KE = constantAt the startPE + KE = mgh + mv2

= 2.0 x 10 x 1.8 + x 2.0 x 52= 36 + 25= 61J

This has all been transformed into PE justbefore it lands.

61J (ANS)

Example 141985 Solution Q6Both projectiles are launched from the sameheight. Therefore they will both take the same

time to fall to the ground.1 (ANS)

Example 151985 Solution Q7To calculate the distance travelledhorizontally, you need to know how long ittook to fall to the ground.Then use d = v x t.Vertically

x= ut + at2u =0

1.8 = x 10 x t2

0.36 = t2t = 0.6 s

Horizontallyd = v x t

d = 5.0 x 0.6= 3.0 m (ANS)

Example 161984 Solution Q6The total energy of Premains constant.

PE + KE = constant

Since P is the same height as the start thePE will be the same. Therefore KEP = KEinitial

E (ANS)

Example 171984 Solution Q7The total energy at the start is equal to thetotal energy at X.InitiallyTotal energy = PE + KE

= MgH + E

At XKE = MgH + E

Mv2= MgH + E

v =2(MgH + E)

M (ANS)

Example 181982 Solution Q5This question just requires you to work outhow long it takes a ball to drop from 5 m tothe ground.

Use Verticallyx= ut + at2

u =0

5.0 = x 10 x t2

1.0 = t2

t = 1 s (ANS)

Example 191982 Solution Q6If the initial horizontal speed was 6.0 m/s then(in 1.0 sec) it will travel

6.0 m (ANS)

Example 201980 Solution Q4KE = mv2

= x 0.2 x 122= 14. 4 J (ANS)

Example 21 1980 Solution Q5

PE = mgh= 0.2 x 10 x 3.0= 6 J (ANS)

Example 22 1980 Solution Q6Initially the ball had 14.4 J of KE, as it rises topoint X it gains 6 J of PE, therefore it mustlose 6 J of KE

KEX= 14. 46.0

KEX= 8.4 J (ANS)

Example 23 1980 Solution Q7KEX= mv

2

8.4 = x 0.2 x v2

v2= 84

v = 9.2 ms

-1

(ANS)


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Example 24 1980 Solution Q8At the point X, the only force acting is theweight. (We are specifically told to neglect airresistance).

Net force (weight) is down.

D (ANS)

DAVID Smith already holds a world record for the greatest distance travelled by a human fired from acannon.

But he added to his list of cannonball achievements yesterday by shooting across the US-Mexico border.

And it was all in the name of art. The feat was the brainchild of a Venezuelan artist, Javier Tellez, and is partof a series of public art projects. Smith climbed into the barrel of the cannon and flashed his US passport asabout 600 people applauded.

He took flight from a popular beach in Tijuana, Mexico, and soared about 45 metres over a line of blackmetal poles. He landed uninjured in a net in Border Field State Park in San Diego, with US Border Patrolagents and an ambulance waiting.

His son, David Smith jnr, is also an accomplished human cannonball. He said his father's flight was the firstacross a border by cannon.

Tellez organised the cannonball launch with psychiatric patients at the Baja California Mental Health Centrein Mexicali, Mexico, as a therapeutic project.

The event is part of an art series that started yesterday, and will run for three months, sponsored by inSite05,a two-nation arts partnership in the San Diego-Tijuana region.

Tellez called the project "living sculpture", and said it was about "dissolving borders" between the US andMexico, and between mental health patients and the rest of the world. "David Smith is a metaphor for flyingover human borders, flying over the law, flying over everything that is established," he said.

Tellez, 36, and Smith snr worked closely on the backdrop, music, costumes and advertising for the project,One Flew Over the Void. Tellez plans a documentary film about it.

Although it is against the law for anyone, including US citizens, to enter the country outside an official port ofentry, Smith was not crossing illegally. Authorities made an exception for him. Smith snr is listed in GuinnessWorld Recordsfor the record distance for a human fired from a cannon.

The Smith family has five cannon-ballers: father, son, two daughters and a cousin. Smith snr built sevencannons designed to fire humans.

The Age 29.08.05

5 projectile motion 2014

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