Projectile motion


A projectile is an object upon which the only force acting is
gravity.
It is assumed that the influence of air resistance is
negligible.
an object
dropped from
rest
an object which is
thrown vertically
upwards
an object is which
thrown upwards at an
angle
1
Monkey and Hunter Experiment
2
Only one force (weight) is acting on the
cannon ball.
Horizontal motion:
Constant velocity
W
W
W
W
W
vertical motion:
W
Constant downward
acceleration g
3
Horizontal motion:
No horizontal force =>
constant velocity
Vertical motion:
Downward force (weight)
=> uniform downward
acceleration (g)
4
A projectile motion
u
Speed of projection: u
Angle of projection: a
Initial horizontal speed: u cos a
u sin a
Initial vertical speed: u sin a
a
u cos a
Horizontal displacement: x = u cos a t
Vertical displacement y: = u sin a t – ½ gt2
5
Time of flight
t=0
y (vertical displacement)
Projectile motion
10
5
0
0
5
10
15
20
25
-5
x (horizontal displacement)




Put y = 0
0 = u sin a t – ½ gt2
t(u sin a – ½ gt) = 0
t = 0 (rejected) or t = 2u sin a / g
t = time of flight
6
Range
y (vertical displacement)
Projectile motion
10
5
0
0
5
10
15
20
25
-5
x (horizontal displacement)
range
Range = final horizontal displacement
= horizontal speed x time of flight
= u cos a x 2u sin a / g = 2 u2 sin a cos a / g
= u2 sin 2 a /g
7
Maximum Range
projectile motion
10
5
0
0
5
10
range15
20
25
Range = 2 u2 sin a cos a / g
= u2 sin 2 a /g
Maximum range = u2 /g when sin 2 a = 1
i.e. 2a = 90o
a = 45o
8
Summary
Vertical motion: y = ut sin q – ½ gt2
 Horizontal motion: x = ut cos q
 Time of flight: Put y = 0
⇒ t = 2usin q / g
 Range: Put t = 2u sin q / g
⇒ x = u2sin 2q / g

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Exercise (HKAL 1998) page 28

A small object is thrown horizontally towards a vertical
wall 1.2 m away. It hits the wall 0.8 m below its initial
horizontal level. At what speed does the object hit the
wall? (Neglect air resistance.)
wall
1.2 m
0.8 m
10
Solution
wall
1.2 m
u


0.8 m
Horizontal motion: x = ut cos q
⇒ 1.2 = ut cos 0o
⇒ ut = 1.2
Vertical motion: y = ut sin q – ½ gt2
⇒ 0.8 = ut sin 0o + ½ gt2
⇒ 0.8 = ½ (10)t2
⇒ 0.8 = ½ (10)t2
⇒ t = 0.4 s
∵ ut = 1.2
∴ u = 1.2/0.4 = 3 ms-1
At the wall, (v = u + at)
Horizontal velocity
= 3 + (0)(0.4) = 3 ms-1
Vertical velocity
= 0 + (10)(0.4) = 4 ms-1
The speed at the wall
= (32 + 42)½ = 5 ms-1
11
Class work page 13
10 ms-1
30 o
100 m

(a) Time of flight:
By s = ut + ½ at2
-100 = 10 sin 30ot + ½ (-10) t2
5t2 – 5t – 100 = 0
(t – 5)(t + 4) = 0
t = 5 s or t = -4 s (rejected)
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Class work page 13
10 ms-1
30 o
100 m

(b) Horizontal distance moved
= u cos 30o x time of flight
= 10 cos 30o x 5
= 43.3 m
13
Class work page 13
10 ms-1
30 o
vx
100 m
q
vy

v
(c) Direction of flight:
vx = 10 cos 30o = 8.66 ms-1
vy = 10 sin 30o + (-10)(5) = -45 ms-1
tan q = |vy / vx| = 45/8.66
q = 79.1o
∴ The particle makes an angle of 79.1o to the
horizontal when it reaches the sea.
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Class work page 13
10 ms-1
30 o
vx
100 m
q
vy

v
(d) Speed of the ball
= sqrt(vx2 + vy2)
= sqrt(8.662 + 452)
= 45.83 ms-1
15
Homework
Textbook page 51 (6, 7)
 Reference book page 116 (8, 12, 16)
 Due date: 2nd October
 Test: Chapter 4 projectile motion
2nd October
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