Page 17 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics
6. Figure 2.7 shows a body projected with a velocity u at an angle θ to the horizontal.

7. Consider the vertical component of motion:
(a) Initial velocity = u sin θ
Acceleration = a = –g
Suppose H = maximum height reached
At the maximum height, vertical component of velocity = 0
2
2
2 Using v = u + 2as
2
0 = (u sin θ) – 2gH
2
2
Maximum height, H = u sin θ
2g
(b) If t 1 = Time taken by the body to reach the maximum height,
using v = u + at
0 = u sin θ – gt 1

t 1 = u sin θ
g

1
(c) The instantaneous height y, at any time t is given by s = ut + at 2
2
1
y = (u sin θ) t – gt 2
2
(d) Let T = Total time of flight, the time taken by the body to travel up and fall back to the ground.
When the body lands on the ground, the vertical displacement s = 0.
1
Using s = ut + at 2
2
1
0 = (u sin θ) T – gT 2
2
2u sin θ
Time of flight, T = g = 2t 1
This means that the time taken by the object to go to its maximum height is the same as the
time it takes to move from the maximum height to the ground.

8. Consider the horizontal component of motion:
(a) Horizontal component of velocity = u cos θ = constant
Instantaneous horizontal displacement at any time t, is
x = (u cos θ) t

(b) The range R of the projectile,
R = (u cos θ) T
g )
= (u cos θ) ( 2u sin θ

2
R = u sin 2θ
g

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02 STPM PHY T1.indd 46 4/9/18 8:19 AM