Crafted by: Alex Ong (Dr)
Revised by: Shige Kudo (Dr)
Reviewed by: Grain Baysa-Pee and Lim Boon Whatt (Dr)
Module Chair: Shige Kudo (Dr)
You are asked to find out an optimum angle of
release (θ) for Randy.
The angle should be θ = 45°?
Or should be 35° < θ < 45°?
Or should be 45° < θ < 55°?
• Kinematics in projectile motion without air resistance
• Kinematics in projectile motion with air resistance
• Objects in free fall
• Theoretical equations and the assumptions
• Back to problem
Kinematics in projectile motion without air resistance
Acceleration
y
ai = initial acceleration
ai
x
Velocity
y
vi = initial velocity
vi
x
Kinematics in projectile motion with air resistance
Acceleration
y
ai = initial acceleration
ai
x
Velocity
y
vi = initial velocity
vi
x
Objects in free fall
Acceleration
g
g
g
g
Weight
(downward
force)
mhouse g
mtiger g
mtiger g
mfeather g
Air resistance
(upward force)
Fair_house
Fair_car
Fair_tiger
Fair_feather
where m = mass, g = 9.81 m/s2, Fair = force due to air drag
Theoretical equation and the assumptions
Height of release is the same height as landing point.
v2 sin 2θ
d=
g
where d = horizontal distance in projectile motion,
v = velocity at the instance of release, θ =release
angle, g = acceleration due to gravity
θ
θ = 45º is optimum, if
• No Fair
• No ball rotation
Theoretical equation and the assumptions
Height of release is different from height as landing point.
v2 sin θ cos θ + v cos θ√ (v sin θ)2 + 2gh
d=
g
θ
h
Assumptions
• No Fair
• No rotation on the ball
where h = height of
release relative to the
height of landing
Back to problem
Height of
release
(m)
Speed of release (m/s)
5
6
• • •
10
• • • •
15
1.5
34° (3.76 m)
37° (4.95 m)
•
41° (11.60 m)
•
43° (11.60 m)
1.6
34° (3.83 m)
36° (5.02 m)
•
41° (11.68 m)
•
43° (11.68 m)
1.7
33° (3.89 m)
36° (5.09 m)
•
41° (11.77 m)
•
43° (11.77 m)
1.8
33° (3.96 m)
35° (5.16 m)
•
41° (11.86 m)
•
43° (11.86 m)
1.9
32° (4.02 m)
35° (5.24 m)
•
41° (11.94 m)
•
43° (11.94 m)
2.0
32° (4.09 m)
35° (5.31 m)
•
40° (12.03 m)
•
43° (12.03 m)
2.1
•
•
•
•
•
42° (24.95 m)
2.2
•
•
•
•
•
42° (25.04 m)
2.3
•
•
•
•
•
42° (25.13 m)
2.4
•
•
•
•
•
42° (25.22 m)
2.5
30° (4.39 m)
33° (5.64 m)
•
39° (12.44 m)
•
42° (25.31 m)
Always θ < 45° using
v2 sin θ cos θ + v cos θ
d=
g
√
(v sin θ)2 + 2gh
Back to problem
• Follow the angle recommended by book, less
than 45°, to maximize the horizontal distance of
the shot .
• However, the release angle may be smaller than
35° because the optimum angle changes due to
the height and speed of release.
What have you learnt?
• Understand the role of initial velocity, angle of release and
height of throw to the range of projectile.
• Understand the influence of gravity on projectile motion.
• Understand the influence of air-drag on projectile trajectory.
• Predict the horizontal distance of projectile motion through
simulation.
Referecnes
• Griffiths, I. W. (2006). Principles of Biomechanics &
Motion Analysis, Lippincott Williams & Wilkins.
• Hay, J. G. (1993). The Biomechanics of Sports
Techniques, Prentice hall.
• Koh, M. &Tah, J. (2006). Understanding Biomechanics
for Physical Education and Sports, Mc Gaw Hill.
• McGinnis, P. M. (2005). Biomechanics of Sport and
Exercise, Human Kinetics.