chapter 2
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Chapter 2 : Kinematics of Linear Motion[ 5 Hours ]
2.1 Linear Motion
2.2 Uniformly accelerated
motion
2.3 Free Falling Body
2.4 Projectile motion
Matriculation Physics SF016 1
2.4 Projectile motion
2.0 Introduction
Kinematics
Description of the motion of objects without consideration of what causes the motion.
2
Learning Outcomes
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to:
•• Define and distinguish between Define and distinguish between
ii) distance and displacement, ) distance and displacement,
ii) speed and velocity, ii) speed and velocity,
iii) instantaneous velocity, average velocity and iii) instantaneous velocity, average velocity and
2.1 Linear Motion
uniform velocityuniform velocity
iv) instantaneous acceleration, average iv) instantaneous acceleration, average
acceleration and uniform accelerationacceleration and uniform acceleration
•• Sketch graphs of displacementSketch graphs of displacement--time, velocitytime, velocity--time and time and accelerationacceleration--time. time.
•• Determine the distance travelled, displacement, velocity Determine the distance travelled, displacement, velocity and acceleration from appropriate graphs.and acceleration from appropriate graphs.
3
Linear motion – motion of an object along a straight line path.
Distance, d -- the total length of travel in moving from one
location to another.-- scalar quantity.-- always positive.
4
-- always positive.
Displacement, s -- straight line distance from the initial
position to the final position of an object.
-- Vector quantity
-- can be positive, negative or zero.
5
Initial Position
final Position
6
Distance travelled = 200mDisplacement = 120 m, in the direction of Northeast
Example
An air plane flies 600 km north and then 400 km to the east.
N E
Final
600 km
400 km
7
Total distance traveled d = 600 + 400 = 1000 km
Displacement, s = = 721.11 km(magnitude)
22 400600 +
Begin
N
600 km
Test your understanding
1. You walk from your house to friend’s house then to the grocery shop. Calculate:
(i) the distance traveled.(ii) the displacement.
8
2. An athlete runs four laps of a 400 m track. What is the athlete’s total displacement?
Speed, v -- Rate of change in distance
-- S.I. unit : m s–1 ; scalar quantity.
t
dv ==
distance that traveltotime
traveleddistance
9
Average Speed, ( )
-- total distance traveled divided by the total
time elapsed in traveling that distance.
t
dvspeedAverage
Σ
Σ=,
v
Velocity, v-- tells us how fast object is moving & in which
direction it is moving.
10
-- is the rate of change in displacement.
-- vector quantity ;
t
svvelocity ==
timetravel
ntdisplaceme,�
SI unit : m s–1
Average velocity, vav
change for the taken time
ntdisplacemein change=avv
�
Instantaneous velocity, v
12
12
tt
ss
t
svav
−
−=
∆
∆=
�
11
Instantaneous velocity, v-- velocity at a specified position or instant of
time along the path of motion.
-- commonly referred as `velocity at point A ` or ` velocity at time t `.
dt
ds
t
sv
t
=∆
∆=
→∆
lim0
-- equal to the gradient at any point on the curve of a displacement – time (s – t) graph.
Slope BE = average velocity
Slope at C = Instantaneous velocity
12
s∆
t∆
Example 1
An insect crawls along the edge of a rectangular swimming pool of length 27m & width 21m. If it crawls from corner A to corner B in 30 min. (a) What is its average speed ?(b) What is the magnitude of its average velocity ?
13
Solution
Given : Length, L = 27m; Width, W = 21m
Total distance traveled, d = 27 + 21 = 48 m
Displacement, s = m21.342127 22=+
dΣ
14
(a) t
dvspeedAverage
Σ
Σ=,
)60(30
48=
10267.0 −= sm
(b)
t
svvelocityAverage av
∆
∆=
�,
)60(30
21.34=
1019.0 −= sm
15
Acceleration, a
-- time rate of change of velocity.
takentime
yin velocit changeonaccelerati =
-- An acceleration may due to:1) change in speed (magnitude), 2) change in direction or3) change in both speed and direction.
-- Velocity is vector quantity, ���� a change in velocity may thus involve either or both magnitude & direction.
16
Quick Test
A car is traveling at 30 km h–1 to the north. Then it turns to the west without changing its speed. Is the car accelerating?
Answer : YES ! Reason : there is a change in direction
-- Deceleration : object is slowing down (direction of acceleration is opposite to the direction of the motion or velocity).
17 18
Car in figure (a) & (d) ���� accelerating
Car in figure (b) & (c) ���� decelerating
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Average acceleration
-- change in velocity divided by the time taken
to make the change.
t
v
tt
vva
∆
∆=
−
−==
12
12
change themake totime
yin velocit change�
-- SI unit : m s–2
-- vector quantity.
20
Instantaneous acceleration
-- acceleration at a particular instant of time.
2
2
0lim
dt
sd
dt
dv
t
va
t
==∆
∆=
→∆
-- SI unit : m s–2
• For linear motion, + and – signs is used to
indicate direction of motion, velocity &
acceleration.
• With horizontal direction we may take :
to the right as +
to the left as –
• With vertical direction we may take :
21
• With vertical direction we may take :
upward as + ; downward as –
The sign convention you choose is entirely up to you. It doesn’t matter as long as you keep the same sign convention for the entire calculation.
Example 2
A car travels in a straight line along a road. Its distance, s is given as a function of time, t by the equation :
32 12.04.2)( ttts −=
(a) Calculate the average velocity of the car for the time interval, t = 0 s and t = 10 s.
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the time interval, t = 0 s and t = 10 s.
(b) Calculate the instantaneous velocity of the car at t = 5 s.
(c) Calculate the instantaneous acceleration of the car at t = 5 s.
Solution
Given : 32 12.04.2)( ttts −=
(a) At t1= 0 s, s1 = 0 m
At t2=10 s,
s2 = 2.4(10)2 – 0.12 (10)3 = 120 m
Average velocity,12
tt
ssvav
−
−=
�
010
0120
−
−=
23
12 ttv
−=
010 −=
112 −= smvav
�
(b) Instantaneous velocity, dt
dsv =
)12.04.2( 32tt
dt
d−=
236.08.4 ttv −=
At t = 5s, 2)5(36.0)5(8.4 −=v
115 −= smv
(c) Instantaneous acceleration,
dva = )36.08.4( 2
ttd
−=
24
dt
dva = )36.08.4( 2
ttdt
d−=
ta 72.08.4 −=
At t = 5s, )5(72.08.4 −=a
22.1 −= sma
Graphical Methods
Displacement – time (s - t) graphs
25
Instantaneous velocity = gradient of (s-t) graphdt
dsv =
velocity – time (v - t) graphs
26
acceleration, a = = gradient of (v - t) graph
Displacement of the object = shaded area
under the (v - t) graph
dt
dv
Example 3 ( PSPM Session 2008/09 )
27
Figure 5 shows a graph of displacement x against
time t of an object moving along x-axis. Calculate
(a) Average velocity for the time interval, 1 s to 4 s.
(b) Average speed for the time interval, 1 s to 4 s.
(c) Instantaneous velocity at t = 2.5 s.
(d) Instantaneous acceleration at t = 5.5 s.
Solution
28
Solution
(a)
t
svaverage
∆
∆=
�
14
2)3(
−
−−=
1s m 67.1 −−=
(b)
t
xvaverage
∆
Σ=
14
342
−
++=
1s m 3 −=
(c) s 2.5 at t gradient |
5.2===
=
dt
dxv
st
23
4)3(
−
−−=
1s m 7 −−=
29
1s m 7 −−=
(d)
dt
dva =� 2s m 0 −
=
Uniform Linear Motion
-- motion with constant velocity
-- acceleration, a = 0 m s–2
The displacement increases by equal amounts in equal times.
30
equal times.
Test your concept
1. Can you accelerate a body without speeding up or slowing down? Is it possible?
2. A car is traveling at 30 km h–1 to the north. Then it turns to the west without changing its speed. Is the car accelerating?
31
3. How would you draw a displacement time graph for a stationary object?
4. What would the gradient of a distance time graph represent?
5. What does the area a speed - time graph represent?
6. The distance – time graph in figure below represents the motion of an ant in 7 seconds. Describe its motion.
32
7. A body moves along the x-axis. Assume that a positive sign represents a direction to the right. The velocity, v of the body is related to time, tthrough the equation
where v and t are measured in and s
232 tv −=
-1s m
33
where v and t are measured in and s respectively. t = 0 when x = 0. Determine(a) the displacement(b) the accelerationat the instant of time t = 1 s
-1s m
Learning Outcomes
At the end of this chapter, students should beAt the end of this chapter, students should be
able to: able to:
•• Derive and apply equations of motion with Derive and apply equations of motion with uniform acceleration:uniform acceleration:
2.2 Linear Motion2.2 Uniformly accelerated motion
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atuv +=
2
2
1atuts +=
asuv 222+=
atuv +=
Uniform acceleration means the accelerationdoes not depend on time or always constant.
velocity changes at a uniform rate.
constant==dt
dva
35
A v - t plot is a straight line whose gradient is equal to acceleration.
Gradient of the slope = a Area under the slope = s
36
Gradient of the slope = a Area under the slope = s
Acceleration, a = constant value
Kinematics Equation for uniform acceleration
Assume a car has uniform acceleration & consider the motion between X and Y :
37
u = initial velocity ( velocity on passing X )v = final velocity ( velocity on passing Y )a = accelerations = displacement ( in moving from X to Y )t = time taken ( to move from X to Y )
velocity – time graph for the car
38
Acceleration, a = gradient of graph v – t
t
uva
−= uvat −=�
rearranged to give:
… (1)tauv +=
Distance traveled, s = area under the graph
= area of trapezium
… (2)tvus )(2
1+=
39
2
Substitute (1) into (2) :
ttauus )(2
1++=
2
2
1tatus += … (3)
From (1) : v = u + at , get an expression for t :
a
uvt
−=
Substitute into (2) :
))((2
1
a
uvuvs
−+=
40
2 a
a
uvuv
2
))(( −+=
222 uvas −=
asuv 222+= … (4)
Example 4
The driver of a pickup truck going 100 km h–1
applies the brakes, giving the truck a uniform
deceleration of 6.50 m s–2 while it travels 20.0m.
(a) What is the speed of the truck in kilometers
per hour at the end of this distance ?
(b) How much time has elapsed ?
41
Solution
Given : a = – 6.50 m s–2 ( deceleration )s = 20.0 mu = 100 km hour –1
1s m78.27s)60(60
m)1000(100 −==
(a) Final velocity , v = ?
asuv 222+=
Convert to km h–1
42
Convert to km h–1
(b) Assume : time elapsed, t
atuv +=
43
Example 5
A park ranger driving on a back country road
suddenly sees a deer ‘frozen’ in his headlights.
The ranger, who is driving at 11.4 m s−1
immediately applies the brakes and slows with
an acceleration of 3.8 m s−2.
44
a) If the deer is 20.0 m from the ranger’s
vehicle when the brakes are applied, how
close does the ranger come to hitting the
deer?
b) How much time is needed for the ranger’s
vehicle to stop?
Solution
45
s
Given: u = 11.4 m s−1 ; a = − 3.80 m s−2
• 1st find the distance traveled before stopped
asuv 222+=From:
The distance between the stopped vehicle & deer:
46
(b) Time needed to stop = ?
atuv +=From:
Example 6
A toy car moves with an acceleration of 2 m s–2
from rest for 2.0 s. It then moves with constant velocity for another 3.0 s. It finally comes to rest after another 1.0 s.
(a) Sketch a velocity-time graph to shown the motion of the toy car.
(b) What is the velocity of the toy car after first
47
(b) What is the velocity of the toy car after first 2 seconds ?
(c) Calculate the deceleration of the car.
(d) What is the total displacement of the car for the whole journey.
(e) Sketch the acceleration-time graph for the motion of toy car.
Solution
(a) v (m s–1)
4
DecelerationDecelerationDecelerationDecelerationV decreases V decreases V decreases V decreases from from from from 4 4 4 4 m sm sm sm s––––1111 to to to to 0 0 0 0 m sm sm sm s––––1111 in in in in 1 1 1 1 ssss
48
(b) Using : atuv +=
t (s)
2 5 60
(c) Using : atuv +=
(d) Displacement, s = area under the graph
49
(d) Displacement, s = area under the graph
(e) Acceleration – time graph
a (m s–2)
2
50
t (s)
2 5 60
2
– 4
Follow up exercise
The speed limit in a school zone is 40 km h–1. A
driver traveling at this speed sees a child run
onto the road 13 m ahead of his car. He applies
the brakes and the car decelerates at a uniform
rate of 8.0 m s–2. If the driver’s reaction time is
0.25 s, will the car stop before hitting the child ?
1
51
0.25 s, will the car stop before hitting the child ?
(a) Is it possible for an object moving at non zero
velocity has a zero acceleration? Explain.
(b) A car is capable of accelerating at 0.60 m s–2.
Calculate the time needed for this car to go from
a speed of 5.5 m s–1 to a speed of 8.0 m s–1.
2
Figure 2 shows a displacement – time graph of a car moving along a straight road.Copy and complete Table 1 by stating any change ( increase / decrease / constant / zero / no change ) in the distance, speed and acceleration of the car for each zone.
3
52
Zone Distance Speed Acceleration
A
B
C
Show that 4 atuv +=
5 Show that 2
2
1atuts +=
6 Show that asuv 222+=
7 An object moves along the x-axis. When it is at the centre of coordinate, its velocity is and its acceleration is . -1s m 6−
-2s m .08
53
Determine(a) its position at t = 2.0 s(b) Its velocity at t = 3.0 s
-2s m .08
8 An object moves along a straight line with constant acceleration. Its initial velocity is After 5.0 s, the velocity becomes Determine the distance travelled during the third second.
.s m 02 -1
.s m 04 -1
Learning Outcomes
At the end of this chapter, students should beAt the end of this chapter, students should be
able to : able to :
•• Describe free falling body.Describe free falling body.
2.3 Free Falling Body
•• Describe free falling body.Describe free falling body.
•• Solve problems on free falling body.Solve problems on free falling body.
54
• Free fall motion is linear vertical motion under the sole influence of gravity.
• The only force acting on the object is the pull of gravity.
• Assumption : free falling
55
• Assumption : free falling objects do not encounter air resistance
• All free falling objects ( on Earth ) always accelerate downwards with an acceleration a = – g
• Value of g = 9.81 m s–2.
• g (vector quantity) is given a minus (–) signindicating that it is always directed downward.
• Replace a with – g into kinematics equation :
Free Fall motion Equationv = u – g tv2 = u2 – 2 g ss = u t – ½ g t2
s = ½ [ u + v ] t
56
s = ½ [ u + v ] twhere
s : vertical displacementu : initial velocityv : final velocityt : time interval g = 9.81 m s–2
• Value of s, u, v may be (+) or (–) depending on the direction of motion.
Upwards Journey:displacement : +velocity : +acceleration : – g
Downwards Journey:Above release point:displacement : +velocity : –acceleration : – gacceleration : – g
Below release point:Displacement : –velocity : –Acceleration : – g
(Reference level/origin)
Graphs of free fall object’s position, velocity & acceleration as functions of time.
Motion graphs for an object thrown vertically upwards and then falling back to the ground.
58
Example 7
A student drops a ball from the top of a tall
building, it takes 2.8 s for the ball to reach the
ground.
(a) What was the ball’s speed just before hitting
the ground ?
(b) What is the height of the building ?
59
(b) What is the height of the building ?
Solution
Given :u = 0 m s–1 (dropped) ; t = 2.8 s ; free fall motion, g = 9.81 m s–2
gtuv −=
)8.2)(81.9(0 −=v
147.27 −−= smv
* ( Minus sign – indicates that v is downward )
1
(a)
60
2
2
1gtuts −=
2)8.2)(81.9(2
10 −=s
downward isnt displaceme : ve-
46.38 ms −=
(b)
Example 8
A boy throws a stone straight upward with an
initial speed of 15 m s–1. What maximum height
will the stone reach before falling back down ?
Solution
At maximum height, object’s velocity is zero for an instant (v = 0 m s–1)
61
gsuv 222−=
s)81.9(2)15()0( 22−=
62.19
225=s m47.11=
Example 9
A stone is thrown vertically downward at an initial speed of 14 m s–1 from a height of 65 m above the ground.(a) How far does the stone travel in 2 s ?(b) What is its velocity just before it hits the
ground ?
Solution :
62
Solution :
Given : u = –14 m s–1 ; g = 9.81 m s–2 ; t = 2 s
(a) Using free fall equation :
2
2
1gtuts −=
(b) Assume the velocity just before hitting the ground = v
gsuv 222−=
63
Example 10
A small pebble is thrown upward from a cliff with an initial velocity 20 m s-1. Calculate(a) Maximum height reached.(b) Time taken to reach a point 25 m below the
initial point.
Solution
64
(a) At max height, v = 0 m s–1
From: gsuv 222−=
65
(b) Given : s = – 25 m ( below initial point )
2
2
1gtuts −=
From:
02520905.4 2=−− tt
From :
a
acbbt
2
42−±−
=
)25)(905.4(4)20()20(2
−−−±−−=
66
)905.4(2
)25)(905.4(4)20()20( −−−±−−=
81.9
84.2920 ±=
only) valueve(1.5 += st
Follow up exercise
1 An object is thrown vertically upwards from a point
on the ground with speed u. Neglect air resistance.
Determine
(a) the maximum height reached by the object
(b) the time taken to return to the starting point in
terms of u and g
67
terms of u and g
2 A ball is thrown vertically upwards from the top of a
building at a speed of 15 m s–1 . If the height of the
building is 200 m, determine
(a) the time taken by the ball to reach the ground
(b) the velocity when the ball reaches the ground
Learning Outcomes
At the end of this chapter, students should beAt the end of this chapter, students should be
able to: able to:
•• Describe projectile motion.Describe projectile motion.
2.4 Projectile Motion
•• Solve problems on projectile motion.Solve problems on projectile motion.
68
• Projectile motion refers to the motion of an object projected into the air at an angle.
69
A motion where object travels at uniform velocityin horizontal direction; at the same time undergoing acceleration in downward directionunder the influence of gravity.
-- 2 dimensional motion.
-- consists of horizontal and vertical motion. These 2 components of motion MUST be discussed separately.
70
-- Assumptions of projectile motion:
1) free fall acceleration, g is constant and is always directed downward.
2) Neglect air resistance.
discussed separately.
(2) Horizontal Projection(1) Projection at an angle
71
• Consider an object thrown with a velocity u at an angle θ° relative to the horizontal.
s ( vertical displacement )
sx ( Horizontal displacement )
72
-- the object move upward or downward it also moving horizontally.
ay= – g
ax= 0 m s – 2
sy ( vertical displacement )
θ
θ
-- The initial velocity, u is resolved into horizontal and vertical components :
Horizontal & vertical motions are
independent and discussed
separately in calculation.
-- the path of motion : parabolic arc
73
θ
θ
sin
cos
uu
uu
y
x
=
=
-- In vertical ( y axis ), gravity acts downwards(ay = – g) ���� object does free fall motion.
-- y-component of the velocity changes with time.
tguv yy −=
• The vertical displacement, sy is given by:
2
2
1tgtus yy −=
and yyy sguv 222−=
• Gravitational force does not act horizontally -- no
ssssyyyy also known also known also known also known as heightas heightas heightas height
74
• Gravitational force does not act horizontally -- no acceleration in horizontal direction ( ax = 0 ), object travels in horizontal direction with uniform velocity.
• Horizontal component of the velocity, vx is constant.
xx uv =
tauv xxx +=From:
0
• With ax = 0, the horizontal displacement, sx is given by:
tus xx =
• The magnitude of the instantaneous velocity v
2
2
1tatus xxx +=
0
75
• The magnitude of the instantaneous velocity vat any point is given by :
22yx vvv +=
x
y
v
v=θtan line horizontal from is θ⇒
Direction of the velocity :
vxvy = 0
For projectile returns to the same vertical level at which it was launched
76
Maximum Height, H
Range, R
• If tH = Time taken by the object to reach the
Using : vy2 = uy
2 – 2g sy
0 = ( u sin θ )2 – 2gH
g
uH
2
sin 22θ
= [ Maximum Height ]
• At the maximum height (H), vy = 0.
77
• If tH = Time taken by the object to reach the
maximum height, H
Using :
vy = uy – g t
0 = ( u sin θ ) – g tH
g
utH
θsin= [ time to reach maximum height ]
Let T = time of flight
When the body lands on the ground, sy = 0
Using :
2
2
1gttus yy −=
2
2
1)sin(0 gTTu −= θ
78
g
uT
θsin2=
2
)sin
(2g
uT
θ= HtT 2=
* As the path is symmetrical, time in going up is
equal to the time in coming down.
⇒
Range, R is the maximum horizontal displacement
traveled.
Horizontal displacement, sx = R when t = T
Using :
sx = uxt
R = u cos θ (T)
)sin2
(cosu
uRθ
θ=
79
)sin2
(cosg
uuR
θθ=
g
u )cossin2(2θθ
=
g
uR
θ2sin2
=
Maximum range, Rmax at a
particular speed is obtained
when :
12sin =θ
°=⇒ 45θ
°= 902θ
80
Object projected horizontally
81
Projectile travels to the right as it falls downward.Initially, we ONLY
have ux , uy = 0
Quick Test
Can you differentiate between free fall and projectile motion ?
82
θ
u
Horizontal displacement sx
83
Vertical displacement,sy
Equations in Projectile Motion
Horizontal (x motion) Vertical (y motion)
θcosuux =
0=xa
θsinuuy =
gay −=
84
tus xx =
xx uv = tguv yy −=
2
2
1tgtus yy −=
yyy sguv 222−=
Procedure for Solving Projectile Procedure for Solving Projectile Procedure for Solving Projectile Procedure for Solving Projectile
Motion Problems Motion Problems Motion Problems Motion Problems
1.1.1.1.Separate the motion into the Separate the motion into the Separate the motion into the Separate the motion into the xxxx(horizontal) part and (horizontal) part and (horizontal) part and (horizontal) part and yyyy (vertical) (vertical) (vertical) (vertical)
part. part. part. part.
85
(horizontal) part and (horizontal) part and (horizontal) part and (horizontal) part and yyyy (vertical) (vertical) (vertical) (vertical)
part. part. part. part.
2. Consider each part separately using 2. Consider each part separately using 2. Consider each part separately using 2. Consider each part separately using
the appropriate equations for x and the appropriate equations for x and the appropriate equations for x and the appropriate equations for x and
y motion.y motion.y motion.y motion.
Example 11
A cannonball is fired with an initial velocity of 30.0 m s–1 at an angle of 35° to the horizontal. (a) What is the maximum height reached by the ball ?(b) What is its range ?
Solution
Given : u = 30.0 m s–1 ; θ = 35° ; ay = g = 9.81 m s–2
86
u= 30 m s–1
θ=35°
R
H
(a) Maximum height ( comp – y ) ; At maximum height, vy = 0
From u = 30 m s–1 & θ = 35˚ , resolved u into x & y comp.
ux = u cos 35° = 30 cos 35= 24.6 m s–1
uy = u sin 35° = 30 sin 35 = 17.2 m s–1
Using :
vy2 = uy
2 – 2gsy
87
vy2 = uy
2 – 2gsy
(0)2 = ( 17.2 )2 – 2 ( 9.81 )(sy)
62.19
84.295=ys m1.15=
Or use equation
g
uH
2
sin 22θ
=)81.9(2
)35(sin)30( 22
= m1.15=
(b) Range = ? Max. horizontal displacement (comp – x)
R = sx = ux t ; to find R, must know value of time of flight, T .
���� Time of flight, T = 2 tup
At max. height , vy = 0
From : vy = uy – gt
(0) = (17.2) – 9.81( tup )
88
(0) = (17.2) – 9.81( tup )
stup 753.181.9
2.17==
Total flight time, T = 2 tup = 2(1.753) = 3.506 s
Range, R = Sx(max) = ux ( T )
= 24.6 (3.506)
= 86.25 m
Example 12
A ball is projected from a height of 25.0 m above the
ground. It is thrown with an initial horizontal velocity of
8.25 m s–1.
(a) How long is the ball in flight before striking the
ground ?
(b) How far from the building does the ball strike the
ground ?
(c) What is the velocity of the ball just before it strikes
89
(c) What is the velocity of the ball just before it strikes
the ground ?
Solution
sx = ?
t = ?
v = ?
(a) From : Sy = uyt – ½ g t 2
( – 25 ) = (0)t – ½ (9.81)(t2)
x-motion y-motion
sx = ? sy = – 25.0 m
ux = 8.25 m/s uy = 0
ax = 0 ay = – 9.81 m/s 2
Consider x and y separately :
90
st 26.281.9
)25(2==
(b) )(tus xx =
m6.18
)26.2)(25.8(
=
=
(c) In horizontal, -- constant velocity motion.
x- component of the velocity is unchanged
vx = ux = 8.25 m s–1
In vertical, g acts on the object, so velocity
changed with time.
From :
vy = uy – g t
= (0) – 9.81( 2.26 )–
xv
v v
θ
91
= (0) – 9.81( 2.26 )
= – 22.17 m s–1
Velocity, v = 22yx vv +
22 )17.22()25.8( −+=166.23 −
= ms
x
y
v
v=θtan
25.8
17.22= °=⇒ 59.69θDirection of v :
v is 23.66 m s–1 at an angle 69.59° below +x axis.
yv v
Example 13
A hockey player hits a “slap shot “ in practice ( with
no goalie present ) when he is 15.0 m directly in
front of the net. The net is 1.20 m high and the puck
is initially hit at an angle of 5° above the ice with a
speed of 35.0 m s–1.
92
speed of 35.0 m s .
(a) Make a sketch of the situation using x - y
coordinates, assuming that the puck is at the
origin at the time it is hit. Be sure to locate the net
in the sketch and show its height.
(b) Determine if the puck makes it into the net.
Solution
(a) Sketching
sy
93
Logical Thinking :
For the puck goes into the net,
���� sy ≤ 1.20 m
initially : θ = 5° ; u = 35.0 m s–1 ; sx = 15 m
(b)
sx= 15 m
From : sx = ux t
ss
tx
43.00.15
===
Resolved u into x & y component:
ux = u cos 5 = 34.9 m s–1
uy = u sin 5 = 3.05 m s–1
In order to find sy, we must calculate t using sx ;
94
su
tx
43.09.34
===
From : sy = uy (t) – ½ g t 2
= (3.05)(0.43) – ½ (9.81)(0.43) 2
= 1.31 – 0.906
= 0.40 m
sy ≤ 1.2 m ���� puck goes into the net.
Exercise (Past Year Question Session 2004/05 Q1)
Figure shows a stationary object on a smooth table at
VB = 1.96 m/s
95
Figure shows a stationary object on a smooth table at height h above the floor. The object moves horizontally a distance of 1.6 m from A to B with uniform acceleration 1.2 m s–2. Then the object is projected from B and fall onto the floor in 0.5 s. Calculate(a) The velocity of the object at B(b) The value of h [ 4 marks ]
Ans : (a) 1.96 ; (b) 1.23
Solution
Motion can be divided into 2 :1.Linear motion on horizontal table A ���� B2.Projectile motion from B & fall on ground
Given : sAB = 1.6 m, a = 1.2 ms-2
time for projectile motion, t = 0.5 s
(a) Using Linear Kinematics Equation
96
asuv 222+=
)6.1)(2.1(2)0( 22+=v
84.32=v
196.1
−= msv
R1
GJU1
(b) h, relate with comp – y. At point B, uy = 0
Using Projectile motion equation:
* Remember : often motion of an object is divided into segments, each with a different acceleration. When solving problems, it is important to realize that the final velocity for one segment is the initial velocity for the next segment.
97
2
2
1tgtus yy −=
2)5.0()81.9(2
1)5.0()0( −=− h
mh 23.1=
R1
GJU1
Mid Term Examination Session 2005/06
A stone is thrown upward from the roof of a building with velocity 15 m s–1 at an angle of 30°to the horizontal. The height of the building is 40.0 m. Calculate
(a) The maximum height of the stone
98
(a) The maximum height of the stone from the ground.
(b) The magnitude of the velocity of the stone just before it strikes the ground.
Answer(a) 42.87 m(b) 31.78 m s–1
Test your understanding
(1) (a) State ONE similarity between free fall and projectile motion.
(b) An airplane moving horizontally with a constant velocity of 115 m s–1 at an altitude of 1050 m. The plane released on aid parcel that falls to the ground.
99
(i) What are the horizontal and vertical components of the parcel’s initial velocity?
(ii) How long does the parcel take to hit the ground?
(iii) Calculate the velocity of the parcel just before it hits the ground.
(2) Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height.
100
Assuming that air resistance is negligible, where will the relief package land relative to the plane?
A. below the plane and behind it. B. directly below the plane.C. below the plane and ahead of it.
(3) A gun with its barrel horizontal fires a shell
from the top of a cliff. Neglecting the effect
of air resistance, which of the three paths A,
B, C most closely resembles the path of the
shell?
101
(4) The diagram shows a trajectory of a golf ball.
Which set of the arrows show the direction of the
acceleration of the ball at point P and Q respectively?
102
A
C
B
D
A transport plane travelling horizontally at a constant velocity of 50 m s−−−−1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculate
a. the flight time of the parcel,
b. the velocity of impact of the parcel,
c. the distance from X to the point of impact.
(Given g = 9.81 m s-2)
(5)
103
(Given g = 9.81 m s-2)
300 m
d
1s m 50 −=u
X
A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure 2.13. If he shoots the ball at a 40.0°°°° angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.
(6)
(Given g = 9.81 m s-2)
104Figure 2.13Figure 2.13
Matriculation Physics SF016 105
End of Chapter 2