Page 18 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics
u 2
(c) The maximum range is g and it occurs when sin 2θ = 1
2θ = 90°
θ = 45°
To obtain the maximum range for a particular speed of projection, the body must be projected
at an angle of 45° to the horizontal.

9. Figure 2.8 shows three possible ranges for a body projected with the same speed but at different 2
angles of projection.

Exam Tips
At the highest point
60° 1. velocity of projectile is
u cos θ horizontally and
45° not zero.
30°
Maximum range 2. kinetic energy
1
= m (u cos θ) 2
The range is maximum when the angle of projection is 45° 2
Figure 2.8

Example 6

A motorcycle stunt-rider moving horizontally takes off from a point 5.0 m above the ground with
a speed of 30 m s . How far away does the motorcycle land?
–1

5.0 m

x
Solution:
Consider the vertical component of motion:
Initial velocity = 0

Acceleration a = +g (Consider downwards direction as positive)
Displacement s = 5.0 m
1
Using s = ut + at 2
2
1
5.0 = 0 + (9.81)t 2
2
 2 × 5.0
Time of flight, t = 9.81 = 1.01 s
Consider horizontal component of motion:
Horizontal displacement x = 30 × t = 30 × 1.01 = 30.3 m

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