MOTION IN 2D:
PROJECTILE MOTION
ERICKA M. DEREZ, LPT
GENERAL PHYSICS 1
RECALL: TYPES OF MOTION
UNIFORM MOTION (UM) UNIFORMLY ACCELERATED
- Constant velocity MOTION (UAM)
- Zero acceleration - Changing velocity
∆𝑥 - Constant acceleration
𝑣𝑎𝑣𝑒 =
∆𝑡 Kinematic Equations (along
horizontal)
Kinematic Equations
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
(along vertical or FREEFALL) 1 2
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡 ∆𝑥 = 𝑣𝑖 𝑡 + 𝑎𝑡
1 2
2 2
2
∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡 𝑣𝑓 = 𝑣𝑖 + 2𝑎∆𝑥
2 2
2
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
PROJECTILE MOTION
• An object may move in both the x and y directions simultaneously
• The form of two dimensional motion we will deal with is called
projectile motion
Examples of projectile Projectile
motion Motion
A cannonball shot from a cannon
a stone thrown into the air
a ball rolling off the edge of a table
a spacecraft circling Earth
PROJECTILE MOTION
With no gravity the
projectile would follow
the straight-line path
(dashed line).
Projectile Motion
But because of gravity it
falls after reaching the
maximum height and
making a curve path
ASSUMPTIONS IN PROJECTILE MOTION
• We may ignore air
friction
• We may ignore the
rotation of the earth Projectile Motion
• With these
assumptions, an object
in projectile motion will
follow a parabolic path
PROJECTILE MOTION
PROJECTILE
TRAJECTORY
Projectile Motion
MAX. HEIGHT, ∆𝑦
RANGE, ∆𝑥
RECALL: COMPONENT METHOD
𝒗𝒊
𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃
Projectile Motion
𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
𝜽
𝐴𝑥 = 𝐴𝑐𝑜𝑠𝜃
𝐴𝑦 = 𝐴𝑠𝑖𝑛𝜃
PROJECTILE MOTION
The x- and y-directions of motion are
completely independent of each other
The x-direction is uniform motion
ax = 0 Projectile Motion
The y-direction is free fall
ay = -g
Each component is
independent of the
other !!!
PROJECTILE MOTION: Horizontal Component
Uniform Motion (UM)
- Constant velocity (𝑣𝑖𝑥 = 𝑣𝑓𝑥 )
- Zero acceleration
Projectile Motion
∆𝑥 𝑅𝑎𝑛𝑔𝑒
𝑣𝑖𝑥 = =
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
ANALOGY: Roll a ball along a horizontal
surface, and its velocity is constant because
no component of gravitational force acts
𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃
horizontally.
PROJECTILE MOTION: Vertical Component
FREE FALL
- Changing velocity 𝑣𝑖𝑦 ≠ 𝑣𝑓𝑦
𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
- Constant acceleration
Projectile Motion 𝑚
(𝑎 = 𝑔 = −9.8 2 )
𝑠
Drop it, and it accelerates downward
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
and covers a greater vertical distance 1 2
∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
each second due to gravitational
2 2
2
force acts 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
VARIATIONS OF PROJECTILE MOTION
1 Full parabola 3 Non-symmetrical
𝑡𝑢𝑝 𝑡𝑢𝑝 ≠ 𝑡𝑑𝑜𝑤𝑛
𝑡𝑑𝑜𝑤𝑛
𝑡𝑡𝑜𝑡𝑎𝑙 = 𝑡𝑢𝑝 + 𝑡𝑑𝑜𝑤𝑛
Projectile Motion
𝑡𝑢𝑝 = 𝑡𝑑𝑜𝑤𝑛 → 𝑡𝑡𝑜𝑡𝑎𝑙 = 2𝑡𝑢𝑝
2 Half-parabola
𝑣𝑖𝑥 An object may be
fired horizontally
𝑡𝑡𝑜𝑡𝑎𝑙 = 𝑡𝑑𝑜𝑤𝑛
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
A. Find the horizontal and vertical components of initial
velocity of golf ball.
B. Determine the maximum height
C. Calculate the time of flight
D. Determine the range
E. Find the velocity at maximum point
F. Velocity after 2.0s.
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
A. Find the horizontal and vertical components of initial
velocity of golf ball.
ILLUSTRATION: GIVEN:
𝑚
𝑣𝑖 = 25.0
𝑠
𝒗𝒊 𝜃 = 30° 𝑓𝑟𝑜𝑚 + 𝑥
𝜽 REQUIRE:
𝑣𝑖𝑥 & 𝑣𝑖𝑦
EQUATIONS IN PROJECTILE MOTION
Uniform Motion (UM) FREE FALL | y - direction
x - direction
1 𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
∆𝑥 𝑅𝑎𝑛𝑔𝑒
1 𝑣𝑖𝑥 = = 2 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
1 2
2 𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 3 ∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
2
2 2
4 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
A. Find the horizontal and vertical components of initial
velocity of golf ball.
ILLUSTRATION: EQUATION & SOLUTION:
𝑚 𝑚
𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 = 25.0 cos 30° = 21.7
𝒗𝒊 𝑠 𝑠
𝑚 𝑚
𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃 = 25.0 sin 30° = 12.5
𝜽 𝑠 𝑠
𝒗𝒊
GIVEN: REQUIRE:
𝑚 𝑣𝑖𝑥 & 𝑣𝑖𝑦 𝒗𝒊𝒚
𝑣𝑖 = 25.0
𝑠 𝜽
𝜃 = 30° 𝑓𝑟𝑜𝑚 + 𝑥 𝒗𝒊𝒙
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
B. Determine the maximum height
ILLUSTRATION:
Max.
height
(∆𝑦)
GIVEN: 𝑚 REQUIRE:
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠
EQUATIONS IN PROJECTILE MOTION
Uniform Motion (UM) FREE FALL | y - direction
x - direction
1 𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
∆𝑥 𝑅𝑎𝑛𝑔𝑒
1 𝑣𝑖𝑥 = = 2 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
1 2
2 𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 3 ∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
2
2 2
4 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
B. Determine the maximum height
ILLUSTRATION: EQUATION & SOLUTION:
2 2 𝑚
Max.
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦 ;𝑔 = −9.8 𝑠2 & 𝑣𝑓𝑦 = 0
2
height
2
−𝑣 𝑖𝑦
(∆𝑦) 0 = 𝑣𝑖𝑦 + 2𝑔∆𝑦 → ∆𝑦 =
2𝑔
𝑚 2
2 − 12.5
GIVEN: 𝑚 REQUIRE: −𝑣𝑖𝑦 𝑠 = 7.97𝑚
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = = 𝑚
𝑣𝑖 = 25.0 𝑠 ∆𝑦 2𝑔 2 −9.8 2
𝑠 𝑚 𝑠
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
C. Calculate the time of flight
ILLUSTRATION:
𝑡𝑢𝑝 𝑡𝑑𝑜𝑤𝑛
GIVEN: 𝑚
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚 REQUIRE:
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠 𝑡𝑡𝑜𝑡𝑎𝑙
EQUATIONS IN PROJECTILE MOTION
Uniform Motion (UM) FREE FALL | y - direction
x - direction
1 𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
∆𝑥 𝑅𝑎𝑛𝑔𝑒
1 𝑣𝑖𝑥 = = 2 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
1 2
2 𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 3 ∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
2
2 2
4 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
C. Calculate the time of flight
ILLUSTRATION: EQUATION & SOLUTION:
𝑚
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡 ;𝑔 = −9.8 𝑠2 & 𝑣𝑓𝑦 = 0
𝑡𝑢𝑝 𝑡𝑑𝑜𝑤𝑛 −𝑣𝑖𝑦
0 = 𝑣𝑖𝑦 + 𝑔𝑡𝑢𝑝 → 𝑡𝑢𝑝 =
𝑔
𝑚
− 12.5 𝑠
GIVEN: 𝑡𝑢𝑝 = 𝑚 ≈ 1.28 𝑠
𝑚 𝑚 −9.8 2
𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚 𝑠
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚 REQUIRE: 𝑡𝑡𝑜𝑡𝑎𝑙 = 𝑡𝑢𝑝 + 𝑡𝑑𝑜𝑤𝑛 = 2𝑡𝑢𝑝
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠 𝑡𝑡𝑜𝑡𝑎𝑙 𝑡𝑡𝑜𝑡𝑎𝑙 = 2 1.28𝑠 = 2.56𝑠
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
D. Determine the range
ILLUSTRATION:
Range (∆𝑥)
GIVEN: 𝑚
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚 𝑡𝑡𝑜𝑡 = 2.56𝑠
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠
EQUATIONS IN PROJECTILE MOTION
Uniform Motion (UM) FREE FALL | y - direction
x - direction
1 𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
∆𝑥 𝑅𝑎𝑛𝑔𝑒
1 𝑣𝑖𝑥 = = 2 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
1 2
2 𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 3 ∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
2
2 2
4 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
D. Determine the range
ILLUSTRATION: REQUIRE:
∆𝑥
EQUATION & SOLUTION:
Range (∆𝑥)
∆𝑥
𝑣𝑖𝑥 = → ∆𝑥 = 𝑣𝑖𝑥 ∆𝑡
GIVEN: ∆𝑡
𝑚 𝑚
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚 ∆𝑥 = 𝑣𝑖𝑥 𝑡𝑡𝑜𝑡 = 21.7 2.56𝑠
𝑣𝑖 = 25.0 𝑠 𝑡 = 2.56𝑠 𝑠
𝑠 𝑚 𝑡𝑜𝑡 ∆𝑥 ≈ 55.6𝑚
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
E. Find the velocity at maximum point
ILLUSTRATION: REQUIRE: 𝑣 at max. pt.
𝑣𝑦 = 0
𝑣𝑥
GIVEN: 𝑚
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚 𝑡𝑡𝑜𝑡 = 2.56𝑠
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠 ∆𝑥 = 55.6𝑚
EQUATIONS IN PROJECTILE MOTION
Uniform Motion (UM) FREE FALL | y - direction
x - direction
1 𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
∆𝑥 𝑅𝑎𝑛𝑔𝑒
1 𝑣𝑖𝑥 = = 2 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
1 2
2 𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 3 ∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
2
2 2
4 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
E. Find the velocity at maximum point
ILLUSTRATION: REQUIRE: 𝑣 at max. pt.
EQUATION & SOLUTION:
Since x-motion is constant velocity
𝑣𝑖𝑥 = 𝑣𝑓𝑥
𝑚
𝑣𝑎𝑡 𝑚𝑎𝑥.𝑝𝑡. = 𝑣𝑖𝑥 = 21.7
GIVEN: 𝑠
𝑚 𝑚
𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚 𝑡𝑡𝑜𝑡 = 2.56𝑠
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠 ∆𝑥 = 55.6𝑚
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
F. Velocity after 2.0s. REQUIRE: 𝑣 at 2.00s
t=1.28s
ILLUSTRATION:
t=2.56s 𝑣𝑥
t=2.00s
𝑣𝑦 𝑣𝑎𝑡 2.00𝑠
GIVEN: 𝑚
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚 EQUATION & SOLUTION:
𝑣𝑖 = 25.0 𝑠 Since x-motion is constant velocity
𝑠 𝑚 𝑡𝑡𝑜𝑡 = 2.56𝑠 𝑚
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠 ∆𝑥 = 55.6𝑚 𝑣𝑥 = 𝑣𝑖𝑥 = 21.7
𝑠
EQUATIONS IN PROJECTILE MOTION
Uniform Motion (UM) FREE FALL | y - direction
x - direction
1 𝑣𝑖𝑦 = 𝒗𝒊 𝑠𝑖𝑛𝜃
∆𝑥 𝑅𝑎𝑛𝑔𝑒
1 𝑣𝑖𝑥 = = 2 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡
∆𝑡 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
1 2
2 𝑣𝑖𝑥 = 𝒗𝒊 𝑐𝑜𝑠𝜃 3 ∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑔𝑡
2
2 2
4 𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 2𝑔∆𝑦
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
F. Velocity after 2.0s. REQUIRE: 𝑣 at 2.00s
t=1.28s
ILLUSTRATION: EQUATION & SOLUTION:
t=2.56s y-motion is freefall
𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑔𝑡 ; 𝑔 = −9.8 𝑚/𝑠 2
t=2.00s
𝑚 𝑚
𝑣𝑓𝑦 = 12.5 + −9.8 2 2.00𝑠
𝑠 𝑠
𝑚
GIVEN: 𝑚 𝑣𝑓𝑦 = −7.1
𝑚 𝑣𝑖𝑥 = 21.7 ∆𝑦 = 7.97𝑚 𝑠
𝑣𝑖 = 25.0 𝑠
𝑠 𝑚 𝑡𝑡𝑜𝑡 = 2.56𝑠
𝜃 = 30° 𝑣𝑖𝑦 = 12.5
𝑠 ∆𝑥 = 55.6𝑚
EXAMPLE #1
A golf ball hits with an initial velocity of 25.0 m/s at an
angle of 30° from the horizontal:
F. Velocity after 2.0s. MAGNITUDE:
𝑚 2 𝑚 2 𝑚
ILLUSTRATION: 𝑣2.00𝑠 = 𝑣𝑥2 + 𝑣𝑦2 = 21.7 + −7.10 ≈ 22.3
𝑠 𝑠 𝑠
ANGLE AND DIRECTION:
t=2.00s 𝑚
𝑣𝑦 −7.1 𝑠 ≈ −18.1°
𝜑 = tan−1 = tan−1 𝑚
𝑣𝑥 21.7 𝑠
EQUATION & SOLUTION:
𝑚 Velocity at 2.00s is:
𝑣𝑥 = 𝑣𝑖𝑥 = 21.7 𝜑
𝑠 𝑣2.00𝑠 = 22.3
𝑚
, 18.1° below the horizontal
𝑚 𝑠
𝑣𝑓𝑦 = −7.1
𝑠
