CHAPTER 2 KINEMATICS OF LINEAR MOTION

3 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.

Zone Distance Speed Acceleration
A
B 52
C

4 Show that v  u  at

5 Show that s  ut  1 at2
2

6 Show that v2  u 2  2as

7 An object moves along the x-axis. When it is at the centre of
coordinate, its velocity is 6 m s-1 and its acceleration is .
8.0 m s-2.

Determine

(a) its position at t = 2.0 s (0, 28) m

(b) its velocity at t = 3.0 s (30 ms-1 )

8 An object moves along a straight line with constant
acceleration. Its initial velocity is 20 m s-1. After 5.0 s, the

velocity becomes 40 m s-1. Determine the distance travelled
during the third second. (78 m)

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2.2.2 Free Falling Body

Learning Outcomes

At the end of this chapter, students should be
able to:
• b) Apply equations of motion for free fall:

a vu v  u  at
t
s  ut  1 at2
s  1 (u  v)t 2
2
v2  u2  2as

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• 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
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.

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• g (vector quantity) is given a minus (–) sign
indicating that it is always directed downward.

• Replace (a) with (–g) into kinematics equation :

Free Fall motion Equation
v=u–gt
v2 = u2 – 2 g s

s = u t – ½ g t2
s=½[u+v]t
where
s : vertical displacement
u : initial velocity
v : final velocity
t : time interval
g = 9.81 m s–2

• Value of s, u, v may be (+) or (–) depending on

the direction of motion. 56

Upwards Journey: Downwards Journey:
displacement : + Above release point:
velocity : + displacement : +
acceleration : – g velocity : –
acceleration : – g
(Reference
level/origin) Below release point:
Displacement : –
velocity : –
Acceleration : – g

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.

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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 ?

Solution
Given :

u = 0 m s–1 (dropped) ;
t = 2.8 s ;
free fall motion, g = 9.81 m s–2

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(a) v  u  gt
v  0  (9.81)(2.8)

v  27.47 m s1

* ( Minus sign – indicates that v is downward )

(b) s  ut  1 gt2
2

s  0  1 (9.81)(2.8)2
2

s  38.46 m
-ve : displacement is below release point.

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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)

v2  u2  2gs
(0)2  (15)2  2(9.81)s

s  225  1 1.4 7 m
19.62

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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 :
Given : u = –14 m s–1 ; g = 9.81 m s–2 ; t = 2 s

(a) Using free fall equation :

s  ut  1 gt2
2

s  (14)(2)  1 (9.81)(2)2
2

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s  28 19.62

s  47.62 m

(b) Assume the velocity just before hitting the
ground = v

v2  u2  2gs

v2  (14)2  2(9.81)(65)
v2  1471.3

v  38.36 ms1

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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: v2  u2  2gs

(0)2  (20)2  2(9.81)s

s  400
19.62

 20 .39 m (ab o v e releas e p o in t )

(b) Given : s = – 25 m ( below initial point )

From: s  ut  1 gt 2
2

 25  20t  1 (9.81)t 2
2

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4.905t 2  20t  25  0

From :

t   b  b2  4ac
2a

 (20)  (20)2  4(4.905)(25)


2(4.905)

 20  29.84
9.81

t  5.1 s ( v e v alu e o n ly )

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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 ~ Derivation of equations~

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
(8.095 s, -64.41 ms-1)

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End of Chapter 2

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