Projectile Motion - Vertical and Horizontal

Projectile motion
Projectile motion is a motion in which an
object is thrown near the earth’s
surface, and it moves along the curved
path under the action of gravity only.

Motion of a snow boarder descending from a slope

Motion of a rider performing a bike stunt

Motion of water jets coming out from nozzles

Assumptions used in projectile motion
1. The effect due to air resistance is negligible.
2. The effect due to curvature of the earth is negligible.
3. The effect due to rotation of the earth is negligible.
4. The acceleration due to gravity is constant over the
range of motion.

Idealized model of projectile motion
θ
v H
R

Projectile
Any object which is projected in the air is called
as projectile.
Projectile
θ
v
H
R

Point of projection
The point from which the object is projected
in air is called as point of projection.
Point of
projection
θ
v
H
R

Velocity of projection
The velocity with which an object is projected
in air is called as velocity of projection.
Velocity of
projection
θ
v
H
R

Angle of projection
The angle with the horizontal at which an object is
projected in air is called as angle of projection.
Angle of
projection
θ
v
H
R

Trajectory
The parabolic path followed by a projectile in air is
called its trajectory.
Trajectory
θ
v
H
R

Time of flight
Time taken by the projectile to cover the entire
trajectory is called as time of flight.
Time of
flight
θ
v
H
R
T

Maximum height of projectile
It is the maximum vertical distance travelled by the
projectile from the ground level during its motion.
Maximum height
θ
v
H
R

Horizontal range of projectile
It is the horizontal distance travelled by the
projectile during entire motion.
Horizontal range
θ
v
H
R

Is horizontal and vertical motions are interdependent?
• Both pink and yellow balls are falling
at the same rate.
• Yellow ball is moving horizontally
while it is falling have no effect on
its vertical motion.
• It means horizontal and vertical
motions are independent of each
other.

Analysis of projectile motion
Motion diagram of a kicked football

vi 𝑥
vi 𝑦
+¿
Projectile motion Vertical motion Horizontal motion
𝑥
𝑦 𝑦
Frame - 1
o o
𝑥
vi
vi 𝑦
vi 𝑥

v1 𝑥
v1 𝑦
+¿
𝑥
𝑦 𝑦
Frame - 2
o o
𝑥
v1 v1 𝑦
v1 𝑥

+¿
𝑥
𝑦 𝑦
v2 𝑦
v2 𝑥
Frame - 3
o o
𝑥
v2

v3 𝑥
v3 𝑦
+¿
𝑥
𝑦 𝑦
v3 𝑦
Frame - 4
o o
𝑥
v3 𝑥
v3

vf 𝑥
vf 𝑦
+¿
𝑥
𝑦 𝑦
Frame - 5
o o 𝑥
vf
vf 𝑥
vf 𝑦

vi 𝑥
vi 𝑦
vi
o
v1 𝑥
v1 𝑦
− g
v2=v2𝑥
v3 𝑥
v3
v3 𝑦
vf
vf 𝑥
vf 𝑦
vi 𝑥 v1 𝑥 v2 𝑥 v3 𝑥
vf 𝑥
v1
vi 𝑦
vf 𝑦
v1 𝑦
v2 𝑦=0
v3 𝑦
𝑦
𝑥

Important conclusion
Projectile motion is a combination of two
motions:
a) Horizontal motion with constant velocity
b)Vertical motion with constant acceleration

Equation of trajectory
The horizontal distance travelled by the
projectile in time ‘t’ along -axis is,
x = v i 𝑥 t
x=( vi cos θ) t
t =
x
vi cos θ
The vertical distance travelled by the projectile
in time ‘t’ along -axis is,
y =v i 𝑦 t −
1
2
g t
2
……..(1)
……..(2)
Substituting the value of ‘t’ from eq.(1) in eq.(2)
we get,
y=(vi s∈θ)t −
1
2
g t
2
y=vi s∈θ ×
x
vi cosθ
−
1
2
g( x
vi cosθ )
2
y =( tan θ) x −
( g
2 vi
2
cos
2
θ )x
2
This is the equation of trajectory of a projectile.
y =α x − β x2
Thus is a quadratic function of . Hence the
trajectory of a projectile is a parabola.
As , and are constant for a given projectile, we
can write
tan θ=¿
α ∧ g
2 vi
2
cos
2
θ
= β ¿

Equation for time of flight
The vertical distance travelled by the
projectile in time ‘t’ along -axis is,
y =v iy t −
1
2
g t
2
For symmetrical parabolic path, time
of ascent is equals to time of
descent.
For entire motion, and
0=(v i s∈θ)T −
1
2
g T
2
1
2
g T 2
=(vi s∈θ)T
T =
2 vi sin θ
g
T A =T D =
T
2
T A=T D=
vi sin θ
g
This is the equation for time of flight.

Equation for horizontal range
The equation of trajectory of projectile is
given by,
y =( tan θ) x −
( g
2 vi
2
cos
2
θ )x
2
This is the equation for horizontal range.
For a given velocity of projection the range
will be maximum when ,
2 θ=90 ° ∧θ=45 °
For entire motion, and
0=( tan θ) R −
( g
2 vi
2
cos
2
θ )R
2
( g
2 vi
2
cos
2
θ )R
2
=(tan θ ) R
g
2 vi
2
cos
2
θ
× R=
sin θ
cosθ
R=
vi
2
( 2sin θ cos θ)
g
R=
vi
2
sin 2 θ
g
Thus range of the projectile is maximum if
it is projected in a direction inclined to the
horizontal at an angle of .

Two angles of projection for the same horizontal range
The equation for horizontal range is
given by,
R=
vi
2
sin 2 θ
g
R ′=
vi
2
sin 2(90 °− θ)
g
Replacing by ,
R ′=
vi
2
sin (180 °− 2 θ)
g
R ′ =
vi
2
sin 2 θ
g
……..(1)
……..(2)
From eq. (1) & (2) we say that,
R = R ′
Thus horizontal range of projectile is same
for any two complementary angles
i.e. and
𝑥
𝑦
o
30°
60°
R=R ′

Equation for maximum height
The third kinematical equation for vertical
motion is,
vf 𝑦
2
= viy
2
−2 g y
At maximum height:
0=( vi sin θ)2
− 2 g H
2 g H =(vi sin θ)2
H =
(vi sin θ )2
2 g
This is the equation for maximum height.
Case-1: Angle of projection is .
H =
(vi sin 45 ° )2
2 g
H =
vi
2
(1/√2)
2
2 g
⇒ H max =
vi
2
4 g
Case-2: Angle of projection is .
H =
(vi sin 90 ° )2
2 g
H =
vi
2
(1)2
2 g
⇒ H =
vi
2
2 g

Sample Problems
Example 1:
A soccer ball is kicked
from the ground at an
angle of 45° to the
horizontal with an initial
velocity of 30 m/s.
Calculate:
a. The time of flight.
b. The maximum height
reached.
c. The range of the
Example 2:
A ball is launched at a
30° angle with the
horizontal at an initial
speed of 25 m/s. Find:
a. The time of flight.
b. The maximum height.
c. The range of the
projectile.

Sample Problems
A football is kicked with an initial velocity
of 50 m/s at an angle of 50° above the
horizontal. Neglecting air resistance,
calculate:
a)The time of flight of the football.
b)The maximum height reached by the
football.
c)The horizontal distance the football
covers before hitting the ground.

Sample Problems
A ball is launched horizontally from a 50-
meter-high cliff with an initial speed of 15
m/s. Neglecting air resistance, answer the
following:
a) How long does it take for the ball to
reach the ground?
b) How far from the base of the cliff will
the ball land?

Horizontal Projectile/Half Projectile
Finding for the time:

Horizontal Projectile
A plane traveling with a horizontal velocity
of 100 m/s is 500 m above the ground. At
some point the pilot decides to drop some
supplies to designated target below.
(a) How long is the drop in the air?
(b)How far away from point where it was
launched will it land?

Horizontal Projectile
A ball is launched horizontally from a 50
meter high cliff with an initial speed of 15
m/s. Neglecting air resistance, answer the
following:
a) How long does it take for the ball to
reach the ground?
b) How far from the base of the cliff will
the ball land?

Projectile Motion - Vertical and Horizontal

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Projectile Motion - Vertical and Horizontal