PHYSICS
CHAPTER 2
CHAPTER 2: Kinematics of linear motion
PHYSICS
CHAPTER 2
Kinematics of linear motion 2.1 2.2 2.3 2.4
Linear Motion Uniformly Accelerated Motion Free Falling Body Projectile Motion
PHYSICS CHAPTER 2 Learning Outcome: 2.1 Linear Motion At the end of this chapter, students should be able to: Define and distinguish between i. ii.
instantaneous velocity, average velocity and uniform velocity. instantaneous acceleration, average acceleration and iv. uniform acceleration. Sketch graphs of displacement-time, velocity-time velocity-time and acceleration-time. Determine the distance travelled, displacement, velocity and acceleration from appropriate graphs. iii.
distance and displacement speed and velocity
PHYSICS
CHAPTER 2
2.1. Linear motion (1-D) 2.1 2. 1.1 .1.. Dis ista tanc nce e,
d
scalar quantity.
is defined as the length of actual path between two points . For example :
Q
P
The length of the path from P to Q is 25 cm.
PHYSICS
CHAPTER 2
2.1.2 Displacement, s s
vector quantity
is defined as the distance between initial point and final point in a straight line. line . The S.I. unit of displacement is metre (m).
Example 1: An object P moves 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 N
relative to the original position. Solution :
O
W
20 m
E
θ
10 m
θ
P
10 m
20 m
PHYSICS
CHAPTER 2
The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v
is defined the rate of change of distance. distance . scalar quantity.
Equation:
speed =
v=
change of distance time interval
Δd Δt
PHYSICS
CHAPTER 2
2.1.4 Velocity, v
is a vector quantity.
The S.I. unit for velocity is m s -1.
Average velocity, vav
is defined as the rate of change of displacement. displacement . Equation: change of displacement
vav
vav
vav
=
=
=
time interval
s2 − s1 t 2
−
t 1
Δs Δt
Its direction is in the same direction of the change in displacement
PHYSICS
CHAPTER 2
Instantaneous velocity velocity,, v
is defined as the instantaneous rate of change of displacement.. displacement Equation:
∆ s v= ∆ t → 0 ∆ t limit
v=
ds dt
An object is moving in uniform velocity if
ds dt
=
constant
PHYSICS
CHAPTER 2
s
s1
Q
The gradient of the tangent to the curve at point Q = the instantaneous velocity at time, t = t 1
0
t
t1 Therefore
s-t graph = velocity Gradient of s-t
PHYSICS
CHAPTER 2
2.1.5 Acceleration, a
vector quantity
The S.I. unit for acceleration is m s -2.
Average acceleration, aav
is defined as the rate of change of velocity. velocity . change of velocity Equation:
aav
=
time interval
− a av = t 2 − v2
aav
=
v1 t 1
Δv Δt
Its direction is in the same direction of motion. motion . The acceleration of an object is uniform when the magnitude of velocity changes at a constant rate and along fixed
PHYSICS
CHAPTER 2
Instantaneous acceleration, a
is defined as the instantaneous rate of change of velocity velocity.. Equation:
∆v a= ∆ t → 0 ∆ t limit
a=
dv dt
=
2
d s 2
dt
An object is moving in uniform acceleration if
dv dt
=
constant
PHYSICS
CHAPTER 2
Deceleration, a
is a negative acceleration. acceleration .
The object is slowing down meaning the speed of the object decreases with time. time .
v
Q
v1
The gradient of the tangent to the curve at point Q = the the instantaneous acceleration at time, t = t 1 0
t1
t
Therefore
Gradient of v-t graph = acceleration acceleration
PHYSICS 2.1.6
CHAPTER 2 Graphical methods
Displacement against time graph ( s-t )
s
s
Gradient increases with time
Gradient = constant
t
0
s (a) Uniform velocity
t
0
(b) The velocity increases with time
Q
(c) P
R
Gradient at point R is negative.
Gradient at point Q is zero.
The direction of velocity is changing.
The velocity is zero.
PHYSICS
CHAPTER 2
Velocity Velocity versus versus time graph ( v-t )
v
v Uniform velocity
v
Uniform acceleration
B
C
A 0
t1 (a) t 2
t
0
t1
(b) t2
t
0
t1
t2 (c)
Area under the v-t graph = displacement
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)
t
PHYSICS
CHAPTER 2 From the equation of instantaneous velocity,
ds
v=
dt
∫ ds = ∫ vdt Therefore
s =
t 2
∫ vdt t 1
s = shaded area under the v − t graph Simulation 2.1
Simulation 2. 2.2
Simulation 2. 2.3
PHYSICS
CHAPTER 2
Example 2 : A toy train moves moves slowly along a straight track according to the displacement, s against time, t graph in figure 2.1.
s (cm) 10 8 6 4 2 Figure 2.1 0
2
4
6
8
10
12
14
t (s) t (s)
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.
PHYSICS
CHAPTER 2
Solution : 0 to 6 s
:
6 to 10 s : 10 to 14 s : b.
v (cm s−1) 1.50
0.68
0
2
4
6
8
10
12
14
t (s) t (s)
PHYSICS
CHAPTER 2
Solution : c. v
d.
av
=
s2 − s1 t 2 − t 1
v = average velocity from 10 s to 14 s s2 − s1 v= t 2 − t 1
PHYSICS
CHAPTER 2
Example 3 : A velocity-time (v-t ) graph in figure 2.2 shows the motion of a lift.
v (m s −1) 4 2 0 -2
5
10
15
20 25
30
35
40
45
50
t (s) t (s)
-4 Figure 2.2
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.
PHYSICS
CHAPTER 2
Solution : a. 0 to 5 s
: Lift moves upward from rest with acceleration of 0.4 m s −2.
5 to 15 s : The velocity of the lift 4 m s−1 but the acceleration
from 2 m s −1 to to 0.2 m s −2.
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 :
20
PHYSICS
CHAPTER 2
Solution : −2 b. a (m s ) 0.8 0.6 0.4 0.2 0 -0.2
5
10
15
20 25 25
30
35
40
45
50
t (s) t (s)
-0.4 -0.6 -0.8
21
PHYSICS
CHAPTER 2
Solution : c. i. v (m s
−1
)
4 2 0 -2
A3
A2
A1 5
10
15
20 25
30 A35 4
40 45 A5
50
t (s) t (s)
-4
Total distance = area under the graph of v-t
= Total distance =
1 2
( 2 ) ( 5) +
1 2
A1 + A 2 + A 3 + A 4 + A 5
( 2 + 4 ) (10 ) +
1 2
( 5 + 10 ) ( 4 ) +
1 2
(5 ) ( 4 ) +
1 2
(15 + 5 ) (4 )
PHYSICS
CHAPTER 2
Solution : c. ii. Displacement
Displacement =
d.
1 2
( 2 ) ( 5) +
aav
1 2
=
area under the graph of v-t
=
A1 + A 2 + A 3 + A 4 + A 5
( 2 + 4) (10 ) +
=
1 2
( 5 + 10 ) ( 4) +
1 2
( 5) ( − 4 ) +
1 2
(15 + 5) ( − 4)
v2 − v1 t 2 − t 1
23
PHYSICS
CHAPTER 2
Exercise 2.1 : 1. Figure Figure 2.3 shows a velocity velocity versus versus time graph graph for for an object constrained to move along a line. The positive direction is to the right.
Figure 2.3
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. s.
PHYSICS
CHAPTER 2
Exercise 2.1 : 2. A train train pulls out out of a station station and accelera accelerates tes steadil steadily y for 20 s until its velocity reaches 8 m s −1. 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 s 2,0 m s 2,-0.267 m s 2, 80 m, 800 m, 120 m; 6.67 m s 1.
PHYSICS CHAPTER 2 Learning Outcome: 2.2 Uniformly accelerated motion At the end of this chapter, students should be able to: Derive and apply equations of motion with uniform acceleration:
v = u + at s = ut +
v
2
=
u
2
+
1
2
at
2 2as
PHYSICS
CHAPTER 2
2.2. Uniformly accelerate accelerated d motion
From the definition of average acceleration, uniform ( constant constant)) acceleration is given by
a=
v− u t
v = u + at where v : final velocity
u : initial velocity a : uniform (constant) acceleration t : time
(1)
PHYSICS
CHAPTER 2 From equation (1), the velocity-time graph is shown in figure velocity 2.4:
v
u Figure 2.4
t
0
time
From the graph, The displacement after time, s = shaded area under the graph = the area of trapezium
Hence,
s =
1 2
( u + v ) t
(2)
PHYSICS
CHAPTER 2 By substituting eq. (1) into eq. (2) thus
s =
1 2
[u + ( u + 1
s = ut +
From eq. (1), From eq. (2),
2
at ) ]t 2
at
(3)
( v − u ) = at
(v + u) =
2 s
multiply
t 2 s
( v + u )( v − u ) =
( at ) t
v
2
=
u
2
+
2as
(4)
PHYSICS
CHAPTER 2 Notes:
equations (1) – (4) can be used if the motion in a straight line with constant acceleration.
For a body moving at constant velocity, ( a = 0) the equations (1) and (4) become
v= u
Therefore the equations (2) and (3) can be written as
s = vt
constant velocity
PHYSICS
CHAPTER 2
Example 4 : A plane on a runway takes 16.2 s over a distance distance of 1200 m to take off from rest. Assuming constant acceleration during take off, calculate a. the speed on leaving the ground, b. the acceleration during take off.
a= ?
Solution :
u= 0
a. Use
s =
1 2
(u +
s = 1200 m t = 16.2 s v ) t
v= ?
PHYSICS
CHAPTER 2
Solution : b. By using the equation of linear motion,
v
2
=
u
2
+
OR
s = ut +
2as
1 2
2
at
PHYSICS
CHAPTER 2
Example 5 : A bus travelling steadily steadily at 30 m s −1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s −2 in the same direction as the bus. Determine a. 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 : a. Given
vb
vc
Use
=
=
30 m s − 1
vb
vc
=
=
constant ; uc
30 m s − 1 u c + ac t c
=
=
0; ac
=
2 ms − 2
PHYSICS
CHAPTER 2
b.
vb
b c
t b
=
uc
=
30 m s − 1
=
b
ac
0
=
0s sc
From the diagram,
t b
uc t c
+
=
1 2
t ; t c sc
ac t c
2
=
vbt b
=
vb
b
2 m s− 2
t b
c
5s
t b
sb
= t − 5 = sb
=
vb
Therefore
sc
=
vb t
=
t
PHYSICS
CHAPTER 2
Example 6 : A particle moves along along horizontal line according to to the equation
s = 3t 3
−
4t 2
+
2t
Where s is displacement in meters and t is time in seconds. At time, t =2.00 s, determine a. the displacement of the particle, b. Its velocity, and c. Its acceleration. Solution : a. t =2.00 s ;
s = 3t 3
−
4t 2
+
2t
PHYSICS
CHAPTER 2
Solution : b. Instantaneous velocity at t = 2.00 s, Use
Thus
v=
ds
v=
d
dt
dt
(3t − 4t + 2t ) 3
2
2 ( ) v = 9 2.00 − 8( 2.00 ) + 2
PHYSICS
CHAPTER 2
Solution : c. Instantaneous acceleration at t = 2.00 s, Use
a=
Hence
dv dt
a = 18( 2.00) − 8
PHYSICS
CHAPTER 2
Exercise 2.2 : 1. A spee speedbo dboat at moving moving at 30.0 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 s
1
2. An unmarked unmarked police police car travellin travelling g a constan constantt 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 car’s 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.
PHYSICS
CHAPTER 2
Exercise 2.2 : 3. A car car trav travel elin ing g 90 90 km 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 s
4. A car driver driver,, travelli travelling ng in his car at a constan constantt velocity velocity of 8 m s-1, sees a dog walking across the road 30 m ahead. The driver’s 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 m
PHYSICS CHAPTER 2 Learning Outcome: 2.3 Free falling body At the end of this chapter, students should be able to: Describe and use equations for free falling body.
For upward and downward motion, use
a = g = 9.81 m s 2
PHYSICS
CHAPTER 2
2.3. Free falling body
is defined as the vertical motion of a body at constant acceleration, g under gravitational field without air resistance. In the earth’s gravitational field, the constant acceleration
known as acceleration due to gravity or free-fall acceleration or gravitational acceleration. acceleration .
the value is g = 9.81 m s
2
the direction is towards the centre of the earth (downward). Note:
In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air resistance.. resistance
PHYSICS
CHAPTER 2 Sign convention:
+ From the sign convention thus,
+
-
a = − g
Table 2.1 shows the equations of linear motion and freely falling bodies. Linear motion
v = u + at v
2
=
s = Table 2.1
u
2
+
2 as 1 2 ut + at 2
Freely falling bodies
v = u − gt
v2
=
s =
u2
−
2 gs 1 2 ut − gt 2
PHYSICS
CHAPTER 2
An example of freely falling body is the motion of a ball thrown thrown vertically upwards with initial velocity, u as shown in figure 2.5. velocity = 0
v= u
H
Figure 2.5
u v
Assuming air resistance is negligible, negligible, the acceleration of the ball, a = − g when the ball moves upward and its velocity decreases to zero when the ball reaches the maximum height, H .
PHYSICS
CHAPTER 2 s
The graphs in figure 2.6 show the motion of the ball moves H up and down.
Derivation of equations At the maximum height height or displacement, H where t = t 1, its velocity,
v= 0 hence
v = u − gt 0 = u − gt 1
therefore the time taken for the ball reaches H ,
Simulation 2.4
t 1
=
u g
v =0
0 v u
t 1
2t 1
0
t 1
2t 1
t 1
2t 1
t
t
− u
a 0
t
PHYSICS
CHAPTER 2 To calculate the maximum height or displacement, H : use either
1
s = ut 1 −
2
gt 12 Where s
OR
v2 = u 2 0 = u2
−
−
2 gs
2 gH
maximum height,
H =
= H
u
2
2 g
Another form of freely falling bodies expressions are
v = u − gt 2 2 = v u − 2 gs s = ut −
1
2
gt 2
v y v y2 s y
= u y − gt = u y2 − 2 gs y 1 2 = u y t − gt 2
PHYSICS
CHAPTER 2
Example 7 : A ball is thrown from the top of a building building is given an initial velocity of 10.0 m s −1 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 B figure 2.7. Calculate a. 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.
u =10.0 m s
d. the velocity of the stone when it reaches point D.
−1
A
(Given g = 9.81 m s −2)
30.0 m
Figure 2.7
C
PHYSICS
CHAPTER 2
Solution : a. At the maximum height, H , v y = 0 and u = u y = 10.0 m s −1 thus
B
v y2
=
u y2
−
2 gs y
u C
A
b. From point A to C, the vertical displacement, s y= 0 m thus
s y 30.0 m
D
=
u y t −
1
2
gt 2
PHYSICS
CHAPTER 2
Solution : c. From point A to D, the vertical displacement, s y= −30.0 m thus
B
s y
=
u y t −
1
2
gt 2
u C
A
a
30.0 m
By using
b
t =
− b ±
c
b
2
2a OR
D
−
4ac Time don’t have negative value.
PHYSICS
CHAPTER 2
Solution : d. Time taken from A to D is t = 3.69 s thus
B
= u y − gt v y = (10.0 ) − ( 9.81) ( 3.69 ) v y
u C
A
OR From A to D, s y = −30.0 m 30.0 m
2
2
= u y − 2 gs y 2 2 v y = (10.0 ) − 2( 9.81) ( − v y
Therefore the ball’s velocity at D is D
30.0 )
PHYSICS
CHAPTER 2
Example 8 : A book is dropped 150 m from the ground. Determine Determine a. 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 :
u y = 0 m s
−1
a. The vertical displacement is
s y = −150 m Hence
s y
= − 150 m
s y 150 m
=
u y t −
1 2
gt 2
PHYSICS
CHAPTER 2
Solution : b. The book’s velocity is given by
u y
s y
=
0
v y
=
u y
− gt
OR
= − 150 m
v y v y
=
2
=
u y
?
Therefore the book’s velocity is
2
−
2 gs y
PHYSICS
CHAPTER 2
Exercise 2.3 : 1. A ball is thrown thrown directly directly downward, downward, with with an an initial initial speed speed of 8.00 m s−1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground, b. the ball’s speed when it reaches the ground. ANS. : 1.79 s; 25.6 m s
1
2. A falling falling stone stone takes takes 0.30 s to travel travel past past a window window 2.2 m tall as shown in figure 2.8.
2.2 m
to travel this distance took 0.30 s
Figure 2.8
From what height above the top of the windows did the stone fall?
PHYSICS CHAPTER 2 Learning Outcome: 2.4 Projectile motion At the end of this chapter, students should be able to: Describe and use equations for projectile,
u x = u y = a x = a y =
u cos θ u sin θ 0 − g
Calculate time of flight, maximum height, range, maximum range, instantaneous position and velocity. velocity.
PHYSICS
CHAPTER 2
2.4. Projectile motion
A projectile motion consists of two components: components:
vertical component (y-comp.)
horizontal component (x-comp.)
motion under constant acceleration, a y= − g motion with constant velocity thus a x= 0
The path followed by a projectile is called trajectory is shown in y figure 2.9.
Simulation 2.5
u y A
v1y
v1
P
θ 1
v1x
u
B
v
s y=H
Q v 2x θ 2
v2y
v2 C
θ
t
t
x
PHYSICS
CHAPTER 2 From figure 2.9,
The x-component of velocity along AC (horizontal) at any point is constant,
u x
=
u cos θ
The y-componen y-componentt (vertical) of velocity varies from one point to another point along AC. but the y-component of the initial velocity is given by
u y
=
u sin θ
PHYSICS
CHAPTER 2 Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.
Velocity
Point P
= v1 y =
v1 x
x-comp. y-comp. magnitude
direction
= u cos θ u y − gt 1
u x
= ( v1 x ) 2 + ( v1 y ) 2 v1 y − 1 θ 1 = tan v1 x v1
Table 2.2
Point Q
= v2 y = v2 x
= u y −
u x
u cos θ
gt 2
= ( v2 x ) 2 + ( v2 y ) 2 v2 y − 1 θ 2 = tan v2 x v2
PHYSICS
CHAPTER 2
2.4.1 Maximum height, H
The ball reaches the highest point at point B at velocity, v where
= v = u x = y-component of the velocity, v y = 0 y-component of the displacement, s y = H 2 2 Use v y = u y − 2 gs y 2 0 = ( u sin θ ) − 2 gH
x-component of the velocity,
H =
2
u sin
2
2 g
θ
v x
u cos θ
PHYSICS
CHAPTER 2
2.4.2 Time taken to reach maximum height, At maximum height, height, H
Time, t
Use
= ∆t ’ and v y= 0
= u y − gt 0 = ( u sin θ ) − g ∆ t '
v y
2.4.3 Flight time,
g
t (from point A to point C)
∆ t = 2∆ t ' ∆ t =
∆ t ' =
u sin θ
2u sin θ g
t’
PHYSICS
CHAPTER 2
R maximum 2.4.4 Horizontal range, R and value of R
Since the x-component for velocity along AC is constant hence
u x
=
v x
=
u cos θ
From the displacement formula with uniform velocity, thus the x-component of displacement along AC is
= u x t and s x = R R = ( u cos θ ) ( ∆ t ) 2u sin θ R = ( u cos θ ) g 2 u ( 2 sin θ cos θ ) R =
s x
g
PHYSICS
CHAPTER 2 From the trigonometry identity,
sin 2θ thus
R =
u
=
2 sin θ cos θ
2
g
sin 2θ
R maximum when The value of R therefore
Rmax
=
u
= =
45 and sin 2 = 1
2
g Simulation 2.6
PHYSICS
CHAPTER 2
2.4.5 Horizontal projectile
Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.
u
u
v x v y
h
Figure 2.10
A
B
x
Horizontal component along path AB.
velocity, u x displacement, s x
v
= u= = x
vx
=
constant
Vertical component along path AB.
initial velocity, u y
displacement, s
=
=
0
Simulation 2.7
h
PHYSICS
CHAPTER 2
Time taken for the ball to reach the floor (point B), t
By using the equation of freely falling bodies,
u y t − gt 2 1 2 h = 0 − gt 2
s y
−
1
=
t =
2
2h g
Horizontal displacement, x
Figure 2.11
Use condition below : 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)
PHYSICS
CHAPTER 2 Since the x-component of velocity along AB is constant, thus the horizontal displacement, x
s x
=
u x t and s x
x = u
=
x
g
2h
Note :
In solving any calculation problem about projectile motion, the air resistance is negligible. negligible .
PHYSICS
CHAPTER 2
Example 9 : y
H
u Figure 2.12 O
θ =
P
60.0° R
v1y Figure 2.12 shows a ball thrown by superman s uperman with an initial speed, u = 200 m s -1 and makes an angle, θ = 60.0° to the horizontal. Determine a. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.
v1x v1 Q
v2y
x v2x v2
PHYSICS
CHAPTER 2
b. the time taken for the ball reaches the maximum maximum height, H and
H . calculate the value of H c. the the hori horizo zont ntal al rang range, e, R d. the magnitude and direction of its velocity when the ball ball reaches the ground (point P). e. the position of the ball, 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 : The component of Initial velocity :
PHYSICS
CHAPTER 2
Solution : a. i. i. positi position on of of the the ball ball when when t = 2.0 2.0 s , Horizontal component :
s x
=
u x t
Vertical component :
s y
=
u y t −
1 2
gt 2
therefore the position of the ball is
PHYSICS
CHAPTER 2
Solution : a. ii. magnitude magnitude and and direction direction of ball’s ball’s velocit velocity y at t = 2.0 s , Horizontal component :
v x
=
u x
Vertical component :
v y
Magnitude,
Direction,
v=
θ = tan
=
= 100 m s − 1
u y
− gt
2 v x
+
v y − 1
2 vy
v = x
= (100 ) + (153)
tan
2
− 1 153
100
2
PHYSICS
CHAPTER 2
Solution : b. i. At the the max maxim imum um heig height ht,, H :
v y
=
0
Thus the time taken to reach maximum height is given by
v y
ii. Apply
s y
=
=
u y
− gt
u y t −
1 2
gt
PHYSICS
CHAPTER 2
Solution : c.
Flight time
= 2×(the time taken to reach the maximum height)
t = 2(17.6 )
Hence the horizontal range, R is
s x
d.
=
u x t
When hen the ball all reach eaches es point oint P thus hus s y The velocity of the ball at point P, Horizontal component: Vertical component:
v1 x
v1 y
=
=
ux
=
0
= 100 m s − 1
u y − gt
PHYSICS
CHAPTER 2
Solution : Magnitude, v1
Direction,
=
2 1 x
v
θ = tan
−1
+
2 1y
v
= (100) + ( − 172)
v1 y = v1 x
2
tan
−1
− 172 100
therefore the direction of ball’s velocity is θ 300 from positive x-axis anticlockwise
=
e.
The tim time tak taken en from from poin pointt O to Q is 45.0 45.0 s. i. positi position on of of the the ball ball when when t = 45.0 45.0 s, Horizontal component :
s x
=
u x t
2
PHYSICS
CHAPTER 2
Solution : Vertical component :
s y
=
u y t −
1 2
gt 2
therefore the position of the ball is (4500 m, 2148 m) e. ii. magnitude magnitude and and direction direction of ball’s ball’s velocit velocity y at t = 45.0 s , Horizontal component :
= u x = 100 m s − 1 Vertical component : v2 y = u y − gt v2 x
PHYSICS
CHAPTER 2
Solution : Magnitude, v2
= v +v 2 v2 = (100 ) + ( −
Direction,
θ = tan
2 2 x
−1
2 2y
269)
2
v2 y v2 x
therefore the direction of ball’s velocity is
PHYSICS
CHAPTER 2
Example 10 : A transport plane travelling travelling at a constant velocity velocity of 50 m s −1 at an altitude of 300 m releases a parcel when w hen 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 o f impact. (given g = 9.81 m s -2) Solution :
u = 50 m s − 1
300 m
X
PHYSICS
CHAPTER 2
Solution : The parcel’s velocity = plane’s velocity
u = 50 m s − 1
thus a.
u x
= u=
50 m s
−1
and
u y
=
0 m s− 1
The vert vertic ical al disp displlacem acemen entt is is giv give en by by Thus the flight time of the parcel is
s y
=
u y t −
1
2
gt 2
PHYSICS
CHAPTER 2
Solution : b. The The com compo pone nent nts s of of vel veloc ocit ity y of of imp impac actt of of the the par parce cel: l: Horizontal component:
v y
Vertical component:
v x
v y
=
v
θ = tan
−1
Magnitude, v
Direction,
2 x
+
=
50 m s − 1 gt
=
ux
=
2 y
= ( 50) + ( −
− = 0 − ( 9.81) ( 7.82 ) v
u y
v y = v x
2
tan
−1
−
76.7 )
76.7 50
therefore the direction of parcel’s velocity is
2
PHYSICS
CHAPTER 2
Solution : c.
Let Let the the dist distan ance ce from from X to the the poi point nt of impa impact ct is d . Thus the distance, d is given by
s x
=
u x t
PHYSICS
CHAPTER 2
Exercise 2.4 : Use gravitational acceleration, g = 9.81 m s −2 1. A basketball basketball player player who is 2.00 2.00 m tall tall is standin standing g on the floor 10.0 m from the basket, as in figure 2.13. 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.
Figure 2.13 ANS. : 10.7 m s
1
PHYSICS
CHAPTER 2
Exercise 2.4 : 2. An appl apple e is thrown thrown at an an angle angle of 30 30 ° above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s−1. Calculate a. 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 the max maxim imum um heig height ht reac reache hed d by by the the appl apple e fro from m the the ground. ANS. : 4.90 s; 170 m; 40.4 m
3. A stone stone is thrown thrown from the top top of one buildin building g toward toward a tall tall building 50 m away. The initial velocity of the ball is 20 m s −1 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.
PHYSICS
CHAPTER 2
THE END… Next Chapter… CHAPTER 3 : Force, Momentum and Impulse
