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PHYSICS CHAPTER 2
CHAPTER 2:CHAPTER 2:Kinematics of linear motionKinematics of linear motion
1
PHYSICS CHAPTER 2
Kinematics of linear motionKinematics of linear motion
2.1 Linear Motion2.2 Uniformly Accelerated Motion2.3 Free Falling Body2.4 Projectile Motion
2
PHYSICS CHAPTER 2
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine and distinguish between and distinguish between
i.i. distance and displacement distance and displacement ii.ii. speed and velocityspeed and velocityiii.iii. instantaneous velocity, average velocity and uniform instantaneous velocity, average velocity and uniform
velocity.velocity.iv.iv. instantaneous acceleration, average acceleration and instantaneous acceleration, average acceleration and
uniform acceleration. uniform acceleration. SketchSketch graphs of displacement-time, velocity-time and graphs of displacement-time, velocity-time and
acceleration-time.acceleration-time. DetermineDetermine the distance travelled, displacement, velocity the distance travelled, displacement, velocity
and acceleration from appropriate graphs.and acceleration from appropriate graphs.
Learning Outcome:2.1 Linear Motion
PHYSICS CHAPTER 2
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2.1. Linear motion (1-D)
2.1.1. Distance, d scalar quantity. is defined as the length of actual path between two pointslength of actual path between two points. For example :
The length of the path from P to Q is 25 cm.
P
Q
PHYSICS CHAPTER 2
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vector quantity is defined as the distance between initial point and final the distance between initial point and final
point in a straight linepoint in a straight line. The S.I. unit of displacement is metre (m).
Example 1:An object P moves 20 m to the east after that 10 m to the south and finally moves 30 m to west. Determine the displacement of P relative to the original position.Solution :Solution :
2.1.2 Displacement,s
N
EW
S
O
P
20 m
10 m
10 m 20 m
PHYSICS CHAPTER 2
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The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v is defined the rate of change of distancerate of change of distance. scalar quantity. Equation:
interval timedistance of changespeed =
tdv =
PHYSICS CHAPTER 2
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is a vector quantity. The S.I. unit for velocity is m s-1.
Average velocity, Average velocity, vvavav is defined as the rate of change of displacementthe rate of change of displacement. Equation:
Its direction is in the same direction of the change in same direction of the change in displacementdisplacement.
2.1.4 Velocity,v
interval timentdisplaceme of change
=avv
tsvav =
12
12av tt
ssv
=
PHYSICS CHAPTER 2
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Instantaneous velocity, Instantaneous velocity, vv is defined as the instantaneous rate of change of the instantaneous rate of change of
displacementdisplacement. Equation:
An object is moving in uniform velocitymoving in uniform velocity if
ts
0tv
=
limit
constant=dtds
dtdsv =
PHYSICS CHAPTER 2
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Therefore
Q
s
t0
s1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous velocity at time, t = t1
Gradient of s-t graph = velocity
PHYSICS CHAPTER 2
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vector quantity The S.I. unit for acceleration is m s-2.
Average acceleration, Average acceleration, aaavav is defined as the rate of change of velocitythe rate of change of velocity. Equation:
Its direction is in the same direction of motionsame direction of motion. The accelerationacceleration of an object is uniformuniform when the magnitude magnitude
of velocity changes at a constant rate and along fixed of velocity changes at a constant rate and along fixed direction.direction.
2.1.5 Acceleration, a
interval time velocityof change
=ava
12
12av tt
vva
=
tvaav =
PHYSICS CHAPTER 2
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Instantaneous acceleration, Instantaneous acceleration, aa is defined as the instantaneous rate of change of velocityinstantaneous rate of change of velocity. Equation:
An object is moving in uniform acceleration moving in uniform acceleration if
tv
0ta
=
limit
constant=dtdv
2
2
dtsd
dtdva ==
PHYSICS CHAPTER 2
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Deceleration,Deceleration, aa is a negative accelerationnegative acceleration. The object is slowing downslowing down meaning the speed of the object speed of the object
decreases with timedecreases with time.
Therefore
v
t
Q
0
v1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous acceleration at time, t = t1
Gradient of v-t graph = acceleration
PHYSICS CHAPTER 2
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Displacement against time graph (Displacement against time graph (s-ts-t))2.1.6 Graphical methods
s
t0
s
t0(a) Uniform velocity (b) The velocity increases with time
Gradient = constant
Gradient increases with time
(c)s
t0
Q
RP
The direction of velocity is changing.
Gradient at point R is negative.
Gradient at point Q is zero.
The velocity is zero.
PHYSICS CHAPTER 2
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Velocity versus time graph (Velocity versus time graph (v-tv-t))
The gradient at point A is positive a > 0(speeding up) The gradient at point B is zero a= 0 The gradient at point C is negative a < 0(slowing down)
t1 t2
v
t0 (a) t2t1
v
t0 (b) t1 t2
v
t0 (c)
Uniform velocityUniform acceleration
Area under the v-t graph = displacement
BC
A
PHYSICS CHAPTER 2
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From the equation of instantaneous velocity,
Therefore
dtdsv =
= vdtds= 2
1
t
tvdts
graph under the area dedsha tvs =
Simulation 2.1 Simulation 2.2 Simulation 2.3
PHYSICS CHAPTER 2
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A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure 2.1.
a. Explain qualitatively the motion of the toy train.b. Sketch a velocity (cm s-1) against time (s) graph.c. Determine the average velocity for the whole journey.d. Calculate the instantaneous velocity at t = 12 s.
Example 2 :
0 2 4 6 8 10 12 14 t (s)
2
4
68
10
s (cm)
Figure 2.1Figure 2.1
PHYSICS CHAPTER 2
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Solution :Solution :0 to 6 s :
6 to 10 s : 10 to 14 s :
b.
0 2 4 6 8 10 12 14 t (s)
0.68
1.50
v (cm s1)
PHYSICS CHAPTER 2
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Solution :Solution :c.
d.
12
12
ttssvav
=
s 14 tos 10 from velocity average=v
12
12
ttssv
=
PHYSICS CHAPTER 2
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A velocity-time (v-t) graph in figure 2.2 shows the motion of a lift.
a. Describe qualitatively the motion of the lift.b. Sketch a graph of acceleration (m s-1) against time (s).c. Determine the total distance travelled by the lift and its displacement.d. Calculate the average acceleration between 20 s to 40 s.
Example 3 :
05 10 15 20 25 30 35 t (s)
-4-2
2
4
v (m s 1)
Figure 2.2Figure 2.2
40 45 50
PHYSICS CHAPTER 2
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Solution :Solution :a. 0 to 5 s : Lift moves upward from rest with
acceleration of 0.4 m s2. 5 to 15 s : The velocity of the lift from 2 m s1 to
4 m s1 but the acceleration to 0.2 m s2. 15 to 20 s : Lift 20 to 25 s : Lift 25 to 30 s : Lift 30 to 35 s : Lift moves
35 to 40 s : Lift moving
40 to 50 s :
PHYSICS CHAPTER 2
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Solution :Solution :b.
t (s)5 10 15 20 25 30 35 40 45 500
-0.4-0.2
0.2
0.6
a (m s2)
-0.6
-0.8
0.8
0.4
PHYSICS CHAPTER 2
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Solution :Solution :c. i.
05 10 15 20 25 30 35 t (s)
-4-2
2
4
v (m s 1)
40 45 50A1
A2 A3
A4 A5
v-t ofgraph under the area distance Total =54321 AAAAA ++++=
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145
214105
211042
2152
21distance Total +++++++=
PHYSICS CHAPTER 2
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Solution :Solution :c. ii.
d.
v-t ofgraph under the areant Displaceme =
54321 AAAAA ++++=
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145
214105
211042
2152
21ntDisplaceme +++++++=
12
12
ttvvaav
=
PHYSICS CHAPTER 2
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Figure 2.3Figure 2.3
1. Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right.
a. Describe the motion of the object in 10 s.b. Sketch a graph of acceleration (m s-2) against time (s) for the whole journey.c. Calculate the displacement of the object in 10 s.
ANS. : 6 mANS. : 6 m
Exercise 2.1 :
PHYSICS CHAPTER 2
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2. A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s1. It then travels at a constant velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s.a. Sketch a velocity-time graph for the journey.b. Calculate the acceleration and the distance travelled in each part of the journey.c. Calculate the average velocity for the journey.Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson Thornes, pg.15, no. 1.11
ANS. : 0.4 m sANS. : 0.4 m s22,0 m s,0 m s22,-0.267 m s,-0.267 m s22, 80 m, 800 m, 120 m; , 80 m, 800 m, 120 m; 6.67 m s6.67 m s11..
Exercise 2.1 :
PHYSICS CHAPTER 2
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Derive and applyDerive and apply equations of motion with uniform equations of motion with uniform
acceleration:acceleration:
Learning Outcome:2.2 Uniformly accelerated motion
atuv +=2
21 atuts +=
asuv 222 +=
PHYSICS CHAPTER 2
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2.2. Uniformly accelerated motion From the definition of average acceleration, uniform (constantconstant)
acceleration is given by
where v : final velocityu : initial velocitya : uniform (constant) accelerationt : time
atuv += (1)t
uva =
PHYSICS CHAPTER 2
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From equation (1), the velocity-time graph is shown in figure 2.4:
From the graph, The displacement after time, s = shaded area under the
graph = the area of trapezium
Hence,
velocity
0
v
utimetFigure 2.4Figure 2.4
( )tvu21s += (2)
PHYSICS CHAPTER 2
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By substituting eq. (1) into eq. (2) thus
From eq. (1),
From eq. (2),
( )[ ]tatuus ++=21
(3)2
21 atuts +=
( ) atuv =( )
tsuv 2=+
multiply
( ) ( ) ( )attsuvuv
=+2
asuv 222 += (4)
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Notes: equations (1) (4) can be used if the motion in a straight motion in a straight
line with constant acceleration.line with constant acceleration. For a body moving at constant velocity, ( ( aa = 0) = 0) the
equations (1) and (4) become
Therefore the equations (2) and (3) can be written asuv =
vts = constant velocityconstant velocity
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A plane on a runway takes 16.2 s over a distance of 1200 m to take off from rest. Assuming constant acceleration during take off, calculatea. the speed on leaving the ground,b. the acceleration during take off.Solution :Solution :
a. Use
Example 4 :
s 2.16=t
?=v
( )tvus +=21
0=u
m 1200=s
?=a
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Solution :Solution :b. By using the equation of linear motion,
asuv 222 +=
OROR2
21 atuts +=
PHYSICS CHAPTER 2
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A bus travelling steadily at 30 m s1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s2 in the same direction as the bus. Determinea. the time taken for the car to acquire the same velocity as the bus,b. the distance travelled by the car when it is level with the bus.Solution :Solution :
a. Given Use
Example 5 :
21 ms 2 0; ;constant s m 30 ==== ccb auv
cccc tauv +=1s m 30 == bc vv
PHYSICS CHAPTER 2
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b.
From the diagram,
c
b
1s m 30 =bv
0=cus 0=bt s 5=bt
2s m 2 =cab
bv b
c
bv
ttb =bc ss =
bc ss =
bbcccc tvtatu =+2
21
Thereforetvs bc =
;ttb = 5= ttc
PHYSICS CHAPTER 2
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A particle moves along horizontal line according to the equation
Where s is displacement in meters and t is time in seconds. At time, t =2.00 s, determinea. the displacement of the particle,b. Its velocity, andc. Its acceleration.Solution :Solution :a. t =2.00 s ;
Example 6 :
ttts 23 243 +=
ttts 23 243 +=
PHYSICS CHAPTER 2
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Solution :Solution :b. Instantaneous velocity at t = 2.00 s,
Use
Thus
dtdsv =
( )tttdtdv 243 23 +=
( ) ( ) 22.0082.009 2 +=v
PHYSICS CHAPTER 2
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Solution :Solution :c. Instantaneous acceleration at t = 2.00 s,
Use
Hence
dtdva =
( ) 82.0018 =a
PHYSICS CHAPTER 2
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1. A speedboat moving at 30.0 m s-1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m s-2 by reducing the throttle.a. How long does it take the boat to reach the buoy?b. What is the velocity of the boat when it reaches the buoy?No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.
ANS. : 4.53 s; 14.1 m sANS. : 4.53 s; 14.1 m s11
2. An unmarked police car travelling a constant 95 km h-1 is passed by a speeder traveling 140 km h-1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police cars acceleration is 2.00 m s-2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.
ANS. : 14.4 sANS. : 14.4 s
Exercise 2.2 :
PHYSICS CHAPTER 2
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3. A car traveling 90 km h-1 is 100 m behind a truck traveling 75 km h-1. Assuming both vehicles moving at constant velocity, calculate the time taken for the car to reach the truck.No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.
ANS. : 24 sANS. : 24 s4. A car driver, travelling in his car at a constant velocity of
8 m s-1, sees a dog walking across the road 30 m ahead. The drivers reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s-2. Calculate the distance from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible.
ANS. : 1.73 mANS. : 1.73 m
Exercise 2.2 :
PHYSICS CHAPTER 2
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Describe and useDescribe and use equations for free falling body. equations for free falling body.
For For upward and downwardupward and downward motion, use motion, useaa = = gg = = 9.81 m s9.81 m s22
Learning Outcome:2.3 Free falling body
PHYSICS CHAPTER 2
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2.3. Free falling body is defined as the vertical motion of a body at constant the vertical motion of a body at constant
acceleration, acceleration, gg under gravitational field under gravitational field without air without air resistanceresistance..
In the earths gravitational field, the constant acceleration known as acceleration due to gravityacceleration due to gravity or free-fall free-fall
accelerationacceleration or gravitational accelerationgravitational acceleration. the value is gg = 9.81 m s= 9.81 m s22 the direction is towards the centre of the earth towards the centre of the earth
(downward).(downward). Note:
In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air assumption made is ignore the air resistanceresistance.
PHYSICS CHAPTER 2
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Sign convention:
Table 2.1 shows the equations of linear motion and freely falling bodies.
Table 2.1Table 2.1
Linear motion Freely falling bodiesatuv += gtuv =
as2uv 22 += gs2uv 22 =2at
21uts += 2gt
21uts =
+
- +
-
From the sign convention thus,
ga =
PHYSICS CHAPTER 2
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An example of freely falling body is the motion of a ball thrown vertically upwards with initial velocity, u as shown in figure 2.5.
Assuming air resistance is negligible, the acceleration of the ball, a = g when the ball moves upward and its velocity velocity decreases to zerodecreases to zero when the ball reaches the maximum maximum height, height, HH.
H
uv
velocity = 0
Figure 2.5Figure 2.5
uv =
PHYSICS CHAPTER 2
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The graphs in figure 2.6 show the motion of the ball moves up and down.
Derivation of equationsDerivation of equations At the maximum height or
displacement, H where t = t1, its velocity,
hence
therefore the time taken for the ball reaches H,
Figure 2.6Figure 2.6
t0
vu
u
t1 2t1
t0
a
g
t1 2t1
t
s
0
H
t1 2t1
v =0
gtuv =1gtu =0
0=v
gut1 =
Simulation 2.4
PHYSICS CHAPTER 2
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To calculate the maximum height or displacement, H:use either
maximum height,
Another form of freely falling bodies expressions are
211 gtuts 2
1=
gsuv 22 2=Where s = H
gHu 20 2 =
OROR
guH2
2
=
gtuv =gsuv 222 =
2
21 gtuts =
gtuv yy =yyy gsuv 2
22=
2
21 gttus yy =
PHYSICS CHAPTER 2
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A ball is thrown from the top of a building is given an initial velocity of 10.0 m s1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in figure 2.7. Calculatea. the maximum height of the stone from point A.b. the time taken from point A to C.c. the time taken from point A to D.d. the velocity of the stone when it reaches point D.(Given g = 9.81 m s2)
Example 7 :
A
B
C
D
u =10.0 m s1
30.0 m
Figure 2.7Figure 2.7
PHYSICS CHAPTER 2
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Solution :Solution :a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus
b. From point A to C, the vertical displacement, sy= 0 m thus
y2y
2y gsuv 2=
2yy gttus 2
1=
A
B
C
D
u
30.0 m
PHYSICS CHAPTER 2
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Solution :Solution :c. From point A to D, the vertical displacement, sy= 30.0 m thus
By using
2yy gttus 2
1=
A
B
C
D
u
30.0 m2a
4acbb 2 =t
ORTime dont Time dont have have negative negative value.value.
a b c
PHYSICS CHAPTER 2
49
Solution :Solution :d. Time taken from A to D is t = 3.69 s thus
From A to D, sy = 30.0 m
Therefore the balls velocity at D is
A
B
C
D
u
30.0 m
gtuv yy =( ) ( ) ( )3.699.8110.0 =yvOR
y2
y2
y gsuv 2=( ) ( )( )30.09.81210.0 = 22yv
PHYSICS CHAPTER 2
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A book is dropped 150 m from the ground. Determinea. the time taken for the book reaches the ground.b. the velocity of the book when it reaches the ground.(given g = 9.81 m s-2)Solution :Solution :
a. The vertical displacement issy = 150 m
Hence
Example 8 :
uy = 0 m s1
150 mm 150=ys
2yy gttus 2
1=
PHYSICS CHAPTER 2
51
Solution :Solution :b. The books velocity is given by
Therefore the books velocity is
gtuv yy =
OR
y2
y2
y gsuv 2=m 150=ys
0=yu
?=yv
PHYSICS CHAPTER 2
52
1. A ball is thrown directly downward, with an initial speed of 8.00 m s1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground,b. the balls speed when it reaches the ground.
ANS. : 1.79 s; 25.6 m sANS. : 1.79 s; 25.6 m s11
2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in figure 2.8.
From what height above the top of the windows did the stone fall?
ANS. : 1.75 mANS. : 1.75 m
Exercise 2.3 :
m 2.2
Figure 2.8Figure 2.8
to travel this distance took 0.30 s
PHYSICS CHAPTER 2
53
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Describe and useDescribe and use equations for projectile, equations for projectile,
CalculateCalculate time of flight, maximum height, range, time of flight, maximum height, range, maximum range, instantaneous position and velocity.maximum range, instantaneous position and velocity.
Learning Outcome:2.4 Projectile motion
uux cos=uu y sin=
0=xagay =
PHYSICS CHAPTER 2
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2.4. Projectile motion A projectile motion consists of two components:
vertical component (y-comp.) motion under constant acceleration, ay= g
horizontal component (x-comp.) motion with constant velocity thus ax= 0
The path followed by a projectile is called trajectory is shown in figure 2.9.
v
u
sx= R
sy=H
ux
v2yuy
v1x
v1y
v2x
v11
v22
t1 t2
B
A
P Q
C
y
xFigure 2.9Figure 2.9
Simulation 2.5
PHYSICS CHAPTER 2
55
From figure 2.9, The x-component of velocityx-component of velocity along AC (horizontal) at any
point is constant,constant,
The y-component (vertical) of velocity variesy-component (vertical) of velocity varies from one point to another point along AC.but the y-component of the initial velocity is given by
uux cos=
uu y sin=
PHYSICS CHAPTER 2
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Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.
Velocity Point P Point Q
x-comp.
y-comp.
magnitude
direction
11 gtuv yy =uuv xx1 cos==
22 gtuv yy =uuv xx2 cos==
( ) ( )2y12x11 vvv +=
=
x1
y111 v
v tan
( ) ( )2y22x22 vvv +=
=
x2
y212 v
v tan
Table 2.2Table 2.2
PHYSICS CHAPTER 2
57
The ball reaches the highest point at point B at velocity, v where x-component of the velocity, y-component of the velocity, y-component of the displacement,
Use
2.4.1 Maximum height, H
uuvv xx cos===0=yv
yyy gsuv 222
=
( ) gHu 2sin0 2 =
guH
2sin 22
=
Hsy =
PHYSICS CHAPTER 2
58
At maximum height, H Time, t = t and vy= 0
Use
2.4.2 Time taken to reach maximum height, t
gtuv yy =( ) 'sin0 tgu =
gut sin' =
2.4.3 Flight time, t (from point A to point C)'2 tt =
gut sin2=
PHYSICS CHAPTER 2
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Since the x-component for velocity along AC is constant hence
From the displacement formula with uniform velocity, thus the x-component of displacement along AC is
2.4.4 Horizontal range, R and value of R maximum
tus xx =
cosuvu xx ==
( ) ( )tuR = cos( )
=
guuR sin2cos
( ) cossin22guR =
and Rsx =
PHYSICS CHAPTER 2
60
From the trigonometry identity,
thus
The value of R maximum when = = 4545 and sin 2sin 2 = = 11 therefore
cossin22sin =
2sin2
guR =
guR
2
max =
Simulation 2.6
PHYSICS CHAPTER 2
61
Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.
Horizontal component along path AB.
Vertical component along path AB.
2.4.5 Horizontal projectile
h
xA B
u u
vxv
yv
Figure 2.10Figure 2.10
constant velocity, === xx vuuxsx = nt,displaceme
0u y = velocity,initialhsy = nt,displaceme
Simulation 2.7
PHYSICS CHAPTER 2
62
Time taken for the ball to reach the floor (point B), Time taken for the ball to reach the floor (point B), tt By using the equation of freely falling bodies,
Horizontal displacement, Horizontal displacement, xx Use condition below :
2yy gttus 2
1=
2gt0h21
=
ght 2=
The time taken for the ball free fall to point A
The time taken for the ball to reach point B=
(Refer to figure 2.11)
Figure 2.11Figure 2.11
PHYSICS CHAPTER 2
63
Since the x-component of velocity along AB is constant, thus the horizontal displacement, x
Note : In solving any calculation problem about projectile motion,
the air resistance is negligibleair resistance is negligible.
tus xx =
=
ghux 2
and xsx =
PHYSICS CHAPTER 2
64
Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s-1 and makes an angle, = 60.0 to the horizontal. Determinea. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.
Example 9 :
Figure 2.12Figure 2.12 xO
u
= 60.0
y
R
H
v2y
v1x
v1y v2xQv1
P
v2
PHYSICS CHAPTER 2
65
b. the time taken for the ball reaches the maximum height, H and calculate the value of H.
c. the horizontal range, Rd. the magnitude and direction of its velocity when the ball
reaches the ground (point P).e. the position of the ball, and the magnitude and direction of its
velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s.
(given g = 9.81 m s-2)Solution :Solution :The component of Initial velocity :
PHYSICS CHAPTER 2
66
Solution :Solution :a. i. position of the ball when t = 2.0 s ,
Horizontal component :
Vertical component :
therefore the position of the ball is
2yy gttus 2
1=
tus xx =
PHYSICS CHAPTER 2
67
Solution :Solution :a. ii. magnitude and direction of balls velocity at t = 2.0 s ,
Horizontal component :
Vertical component :
Magnitude,
Direction,
gtuv yy =
1xx uv
== s m 100
( ) ( ) 2222 153100 +=+= yx vvv
=
=
100153tantan 11
x
y
vv
PHYSICS CHAPTER 2
68
Solution :Solution :b. i. At the maximum height, H :
Thus the time taken to reach maximum height is given by
ii. Apply
gtuv yy =
0=yv
gttus yy 21
=
PHYSICS CHAPTER 2
69
Solution :Solution :c. Flight time = 2(the time taken to reach the maximum height)
Hence the horizontal range, R is
d. When the ball reaches point P thusThe velocity of the ball at point P,Horizontal component:Vertical component:
( )17.62=t
tus xx =
11 s m 100
== xx uv
0=ys
gtuv yy =1
PHYSICS CHAPTER 2
70
Solution :Solution :Magnitude,
Direction,
therefore the direction of balls velocity is
e. The time taken from point O to Q is 45.0 s. i. position of the ball when t = 45.0 s,
Horizontal component :
( ) ( ) 2221211 172100 +=+= yx vvv
=
=
100172tantan 1
1
11
x
y
vv
300= from positive x-axis anticlockwisefrom positive x-axis anticlockwise
tus xx =
PHYSICS CHAPTER 2
71
Solution :Solution :Vertical component :
therefore the position of the ball is (4500 m, (4500 m, 2148 m)2148 m)e. ii. magnitude and direction of balls velocity at t = 45.0 s ,
Horizontal component :
Vertical component :
2yy gttus 2
1=
gtuv yy =2
12 s m 100
== xx uv
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Solution :Solution :Magnitude,
Direction,
therefore the direction of balls velocity is
( ) ( ) 222 269100 +=v22
222 yx vvv +=
=
x
y
vv
2
21tan
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A transport plane travelling at a constant velocity of 50 m s1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculatea. 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)Solution :Solution :
Example 10 :
300 m
d
1s m 50 =u
X
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Solution :Solution :The parcels velocity = planes velocity
thus
a. The vertical displacement is given by
Thus the flight time of the parcel is
1s m 50 == uux
1s m 50 =uand 1s m 0 =yu
2
21 gttus yy =
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Solution :Solution :b. The components of velocity of impact of the parcel:
Horizontal component:Vertical component:
Magnitude,
Direction,
therefore the direction of parcels velocity is
1s m 50 == xx uv
( ) ( )7.829.810 =yvgtuv yy =
=
=
506.77tantan 11
x
y
vv
( ) ( ) 2222 6.7750 +=+= yx vvv
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Solution :Solution :c. Let the distance from X to the point of impact is d.
Thus the distance, d is given by
tus xx =
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Figure 2.13Figure 2.13
Use gravitational acceleration, g = 9.81 m s21. 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.
ANS. : 10.7 m sANS. : 10.7 m s11
Exercise 2.4 :
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2. An apple is thrown at an angle of 30 above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s1. Calculatea. the time taken for the apple to strikes the ground,b. the distance from the foot of the building will it strikes
the ground,c. the maximum height reached by the apple from the
ground.ANS. : 4.90 s; 170 m; 40.4 mANS. : 4.90 s; 170 m; 40.4 m
3. A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s1 at 40 above the horizontal. How far above or below its original level will the stone strike the opposite wall?
ANS. : 10.3 m below the original level.ANS. : 10.3 m below the original level.
Exercise 2.4 :
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THE ENDNext Chapter
CHAPTER 3 :Force, Momentum and Impulse
Slide 1Kinematics of linear motion Learning Outcome:2.1. Linear motion (1-D)Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 262.2. Uniformly accelerated motionSlide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 402.3. Free falling bodySlide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 532.4. Projectile motionSlide 55Slide 56Slide 57Slide 58Slide 59Slide 60Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Slide 67Slide 68Slide 69Slide 70Slide 71Slide 72Slide 73Slide 74Slide 75Slide 76Slide 77Slide 78Slide 79