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Linear motion

When the body moves upwards (see (b) To fi nd time taken to reach
Figure 6.20 (b)), the equations will change maximum height, use;
to:

(i) v = g u - t v u= - gt
1 2
(ii) h ut= - gt 2 0 20 m/s= - 1 m/s0 t ´
2
(iii) v = u - 2gh t = 20 m/s
2
2
10 m/s 2
The negative sign in the equations
indicates that the body is moving upwards = 2 s
against acceleration due to gravity.
So, the body initially starts with 20 m/s
For a body moving upwards, the velocity and used 2 s to reach a height of 20 m.
at the turning point is zero.
Exercise 8.4
Example 8.10
1. A stone falling down a well takes 2 s
A body moved vertically upward to a to reach the water surface. Calculate:
maximum height of 20 m. Calculate: (a) the velocity with which the

(a) the initial velocity. stone hits the water surface.
(b) the distance of the water surface
(b) the time taken to reach the from the top of the well shaft.
maximum height.
2. A stone is thrown vertically upwards
Solution with an initial velocity of 29.4 m/s
from the top of a tower 34.3 m high.
(a) From v = 2 2 u - ;gh Find:
2
where (a) the time taken to reach the
v = 0 m/s, h = 20 m and g = 10 m/s maximum height.
2
(b) the total time that elapses just
0 =v v 2 u = 20 m/su 2 2 2 -´2 10m/s 20m-´ 2 ´ t ´ before it reaches the ground.
2 10m/s 20m´1 m/s0- =
3. A car is travelling at 20 m/s along a
2
2 10 m/s 20m´ m/s0-
0 =v 0 u=20 m/su = 2 2 -´12 10m/s 20m-´ 2 ´ t ´ straight road. The brakes are applied
for 5 s causing a retardation of
2
0 =v 2 u =m/s - 2 2 -´ 010m/s´ t 20m 3m/s . Find the car’s fi nal velocity.
´1 m/s ´ 2
2
2 10 m/s 20 m´ 20u =
Determination of the acceleration due
2
u = 400 m /s 2
2
to gravity
u = 20 m/s Early studies related to motion of

Therefore, the initial velocity is objects under gravity were conducted
th
20 m/s. by Galileo Galilei between the 16 and
17 centuries. Galileo was an Italian
th
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Physics Form 1 Final.indd 187 16/10/2024 20:58