Page 196 - Physics

Physics for Secondary Schools

Note: The accepted value of g is Hence,
9.81 m/s . 0m/s =50m/s 10m/s– 2 ´ t
2
T 2
The values of are constant, showing 10m/s t = 50m/s
2
l
that T is directly proportional to l, t = 50 m/s
2
10 m/s 2
That is, T 2 l µ
t = 5 s.
T 2 4p 2
Thus, = = constant It will take 5 seconds to reach the
l g maximum height.

4 l (b) From the second equation of
2
This implies that g =
T 2 motion:
The graph obtained was a straight line s = ut + 1 at 2
2
passing through the origin. 1
The expected value for gravitational h = ut + 2 gt 2
acceleration is 9.81 m/s . When a body is
2

moving against gravity, g – 9.81 m/s . h = 125 m
2
=
However, the value of g is usually = 250m – 125 m
approximated to 10 m/s . = 125 m
2

Example 8.11 The maximum height reached is
125 m.
An object is thrown vertically upwards
with an initial velocity of 50 m/s. (c) When the object returns to its
(a) How long will it take to reach its starting point its displacement is
maximum height? zero (s = 0).

(b) To what height will it rise? From the third equation of motion
(c) What will be its velocity when it
returns to its starting point? where, u = 50 m/s
(d) How long will it be in the air? a = g = –10 m/s 2
Hence, v = 2 u + 2 2 gs
Solution 2 2
(a) At its maximum height, the v = u + 0
2
velocity of thr object is zero (v = 0). v = u 2
From the fi rst equation of motion: v = u 2
v u at=+

But, a = g = –10 m/s 2 The object has a velocity of 50 m/s
u = 50 m/s on its return.

190
Student’s Book Form One

Physics Form 1 Final.indd 190 16/10/2024 20:58