Vertical Two-Dimensional Motion
Projectile Motion: Horizontal Launch
A projectile follows a curved path, because it
has a combination of horizontal (x-axis) and
vertical (y-axis) velocity vectors:
The resultant of
horizontal and vertical
components is a
diagonal vector that
changes as the object
falls
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Horizontal Launch
Projectile Motion: Other Angles
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Specific Outcomes:
i. I can explain, quantitatively, two-dimensional motion in a horizontal
or vertical plane, using vector components.
ii. I can solve, quantitatively, projectile motion problems near Earth’s
surface, ignoring air resistance.
Vertical Two-Dimensional Motion
Projectile Motion: Horizontal Launch
Horizontal and vertical velocities are
separate, but have the same time, t
This is because the object is moving
horizontally and vertically at the same time
Initially:
• horizontally: vx (remains constant)
• vertically: vi = 0 m/s
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
1
Projectile Motion: Horizontal Launch
After time, t:
• horizontally: vx is still constant
• vertically: magnitude of vf increases (due
to gravity)
Projectile Motion: Horizontal Launch
Horizontally, we will use this equation where
d is the range of the object:
d = vavet
Vertically, we will use the following:
vf - vi
Use the time (t) to do calculations between
horizontal and vertical velocities
a=
We will use our kinematics equations here
d = vit + ½at2
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
t
d=
( )
vi + vf
2
t
vf2 = vi2 + 2ad
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Horizontal Launch
Projectile Motion: Horizontal Launch
ex. 1: A stone is thrown horizontally at 15 m/s
East from a cliff top 44 m high.
ex. 1: A stone is thrown horizontally at 15 m/s
East from a cliff top 44 m high.
a) How long does it take to reach the
bottom of the cliff?
d = vit + ½at2
t=
t=
2(-44 m) – 2(0 m/s)t
(-9.81 m/s2)
2d – 2vit
a
= 2.995…s
= 3.0 s
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
b) How far from the base of the cliff
does the stone strike the ground?
horizontal velocity = CONSTANT
d = vavet = (+15 m/s)(2.995…s)
= 45 m [E]
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
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Projectile Motion: Horizontal Launch
Projectile Motion: Horizontal Launch
ex. 1: A stone is thrown horizontally at 15 m/s
East from a cliff top 44 m high.
ex. 2: An object is thrown horizontally at a
velocity of 10.0 m/s South from the top
of a 90.0 m high building.
c) What is the vertical velocity as the
stone hits the ground?
= 29 m/s [down]
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
a) How long does it take to hit the
ground?
2d – 2vit
t=
d = vit + ½at2
a
t=
2(-90.0 m) – 2(0 m/s)t
= 4.283…s
(-9.81 m/s2)
= 4.28 s
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Horizontal Launch
Projectile Motion: Horizontal Launch
ex. 2: An object is thrown horizontally at a
velocity of 10.0 m/s South from the top
of a 90.0 m high building.
ex. 3: An object is thrown horizontally at 20.0
m/s from the top of a cliff. The object
hits the ground 48.0 m from the base of
the cliff:
b) What is the range of the object?
horizontal velocity = CONSTANT
a) How high is the cliff?
horizontally,
d = vavet = (-10.0 m/s)(4.283…s)
= -42.8 m = 42.8 m [S]
d = vavet
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
t=
d
48.0 m
=
= 2.40 s
v
20.0 m/s
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Projectile Motion: Horizontal Launch
Projectile Motion: Horizontal Launch
ex. 3: An object is thrown horizontally at 20.0
m/s from the top of a cliff. The object
hits the ground 48.0 m from the base of
the cliff:
ex. 3: An object is thrown horizontally at 20.0
m/s from the top of a cliff. The object
hits the ground 48.0 m from the base of
the cliff:
a) How high is the cliff?
as a scalar, d = vit + ½at2
d = (0 m/s)(2.40 s) + ½(-9.81 m/s2)(2.40 s)2
= -28.3 m = 28.3 m
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Other Angles
A projectile may also be launched at an
angle:
b) What is the vertical velocity as the
vf - vi
object hits the ground?
a=
vf = vi + at = 0 m/s + (-9.81
m/s2)(2.40
s)
t
= -23.5 m/s = 23.5 m/s [down]
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Other Angles
Horizontal and vertical velocities are still
separate and still have the same time, t
Horizontally, vx remains constant throughout
Initially: vertical vi is positive (going up)
After time, t, the magnitude of vf:
• decreases if the object is going up
• increases if the object is going down
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
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Projectile Motion: Other Angles
Since we are launching at an angle, we must
separate our original velocity into its x-axis
and y-axis components
Projectile Motion: Other Angles
Initial velocity may be visualized with x and y
components:
trajectory
We will use trigonometry to do this
vy
Remember that an object will gain height
until its vertical velocity becomes zero
θ
vx
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Other Angles
To determine x-axis and y-axis component
velocities, use SOHCAHTOA:
HYP
θ
OPP
sinθ =
HYP
OPP
ex.
A cricket ball is launched at an initial
velocity of 30 m/s [35°] from the ground,
in a northern direction.
a) Calculate the horizontal velocity.
ADJ
ADJ
cosθ =
HYP
Projectile Motion: Other Angles
30 m/s
OPP
tanθ =
ADJ
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
vy
35°
vx
vx = (30 m/s) cos35 = 24.5…m/s
= 25 m/s [N]
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
5
Projectile Motion: Other Angles
ex.
A cricket ball is launched at an initial
velocity of 30 m/s [35°] from the ground,
in a northern direction.
Projectile Motion: Other Angles
ex.
b) Determine the initial vertical velocity
vy = (30 m/s) sin35 = 17.2…m/s
c) How long does it take for the ball to
reach maximum height?
vf - vi
vf - vi
a=
t=
t
a
(0 m/s) – (+17.2…m/s)
t=
= 1.7 s
-9.81 m/s2
= 17 m/s [up]
= vi
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Projectile Motion: Other Angles
ex.
A cricket ball is launched at an initial
velocity of 30 m/s [35°] from the ground,
in a northern direction.
d) Calculate the total time spent by the
ball in the air.
vf - vi
assuming uniform acceleration, t =
a
(-17.2…m/s) – (+17.2…m/s)
t=
-9.81 m/s2
A cricket ball is launched at an initial
velocity of 30 m/s [35°] from the ground,
in a northern direction.
Projectile Motion: Other Angles
ex.
A cricket ball is launched at an initial
velocity of 30 m/s [35°] from the ground,
in a northern direction.
e) Determine the range of the cricket
ball.
dx = vxt = (24.5…m/s)(3.46…s)
= 87 m [N]
= 3.46…s = 3.5 s
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
Dulku – Physics 20 – Unit 1 (Kinematics) – Topic L
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