Page 12 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics

Example 5

(a) A body accelerates uniformly from rest along a straight line. Sketch a graph to show the
variation of displacement with time. How can the instantaneous velocity be deduced from the
displacement-time graph?
(b) A cricket player throws a ball vertically upwards and catches it 3.0 s later. Neglecting air
resistance, calculate 2
(i) speed of the ball when it leaves the player’s hand,
(ii) the maximum height reached by the ball.
(c) Sketch a graph to show how the velocity of the ball varies with time. Mark each of the following
instants on the graph:
(i) The ball leaves the player’s hand (t 1 ).
(ii) The ball at the maximum height (t 2 ).
(iii) The ball returns to the player’s hand again (t 3 ).

(d) You were told that the air resistance is negligible. In actual fact the ball experiences a retarding
force during its motion. Without making any calculation, explain how the air resistance affects
(i) the time taken to reach the highest point,
(ii) the value of the maximum height reached.

(e) Discuss, taking into account the air resistance, whether the time t u taken by the ball to reach
the highest point is greater or smaller than t d , the time taken for the ball to drop.
–2
(Assume g = 10 m s ).
Solution:

(a) Instantaneous velocity = Gradient of the graph at time, t
s

0
t
Note that the gradient of the graph = 0 when t = 0
because initial velocity = 0

(b) (i) Suppose u = velocity of ball when it leaves (ii) At the maximum height H,
the player’s hand. velocity v = 0

Acceleration a = –g Using v = u + 2as
2
2
= –10 m s –2 0 = 15 – 2 × 10 × H
2
Time t = 3.0 s H = 11.25 m
Displacement s = 0
1
Using s = ut + at 2
2
1
0 = 3.0 u – × 10 × (3.0) 2
2
u = 15 m s –1

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