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MR. UMAKANT BHASKARRAO GOHATRE
Projectile Motion
What is projectile?
Projectile -Any object which projected by some
means and continues to move due to its own inertia
(mass).
Projectiles move in TWO dimensions
Since a projectile
moves in 2-
dimensions, it
therefore has 2
components just
like a resultant
vector.
Horizontal and
Vertical
Horizontal “Velocity” Component
NEVER changes, covers equal displacements in
equal time periods. This means the initial
horizontal velocity equals the final horizontal
velocity
In other words, the horizontal
velocity is CONSTANT. BUT
WHY?
Gravity DOES NOT work
horizontally to increase or
decrease the velocity.
Vertical “Velocity” Component
Changes (due to gravity), does NOT cover equal
displacements in equal time periods.
Both the MAGNITUDE and DIRECTION change. As
the projectile moves up the MAGNITUDE
DECREASES and its direction is UPWARD. As it
moves down the MAGNITUDE INCREASES and the
direction is DOWNWARD.
Combining the Components
Together, these
components produce
what is called a
trajectory or path.
This path is parabolic
in nature.
Component Magnitude Direction
Horizontal Constant Constant
Vertical Changes Changes
Horizontally Launched Projectiles
Projectiles which have NO upward trajectory and NO initial
VERTICAL velocity.
0 /oyv m s=
constantox xv v= =
Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for the
“y” direction. And for this we use kinematic #2.
21
2oxx v t at= +
oxx v t=
Remember, the velocity is
CONSTANT horizontally, so
that means the acceleration
is ZERO!
21
2
y gt=
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is
equal to ZERO.
Horizontally Launched Projectiles
Example: 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?
What do I
know?
What I want to
know?
vox=100 m/s t = ?
y = 500 m x = ?
voy= 0 m/s
g = -9.8 m/s/s
2 2
2
1 1500 ( 9.8)
2 2
102.04
y gt t
t t
= → − = −
= → = 10.1 seconds
(100)(10.1)oxx v t= = = 1010 m
Vertically Launched Projectiles
Component Magnitude Direction
Horizontal Constant Constant
Vertical Decreases up, 0
@ top, Increases
down
Changes
Horizontal Velocity
is constant
Vertical
Velocity
decreases
on the way
upward
Vertical Velocity
increases on the
way down,
NO Vertical Velocity at the top of the trajectory.
Vertically Launched Projectiles
Since the projectile was launched at a angle, the
velocity MUST be broken into components!!!
cos
sin
ox o
oy o
v v
v v
θ
θ
=
=vo
vox
voy
θ
Vertically Launched Projectiles
There are several things
you must consider
when doing these types
of projectiles besides
using components. If it
begins and ends at
ground level, the “y”
displacement is ZERO:
y = 0
Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
cos
sin
ox o
oy o
v v
v v
θ
θ
=
=
vo
vox
voy
θ
oxx v t= 21
2oyy v t gt= +
Example
A place kicker kicks a football with a velocity of 20.0
m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
cos
20cos53 12.04 /
sin
20sin53 15.97 /
ox o
ox
oy o
oy
v v
v m s
v v
v m s
θ
θ
=
= =
=
= =
v o
=20.0
m
/s
θ = 53
Example
A place kicker kicks a
football with a
velocity of 20.0 m/s
and at an angle of 53
degrees.
(a) How long is the
ball in the air?
What I know What I want
to know
vox=12.04 m/s t = ?
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8
m/s/s
2 2
2
1 0 (15.97) 4.9
2
15.97 4.9 15.97 4.9
oyy v t gt t t
t t t
t
= + → = −
− = − → =
= 3.26 s
Example
A place kicker kicks a
football with a
velocity of 20.0 m/s
and at an angle of 53
degrees.
(b) How far away does
it land?
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = ?
y = 0 ymax=?
g = - 9.8
m/s/s
(12.04)(3.26)oxx v t= → = 39.24 m
Example
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(c) How high does it
travel?
CUT YOUR TIME IN
HALF!
What I know What I want
to know
vox=12.04 m/s t = 3.26 s
voy=15.97 m/s x = 39.24 m
y = 0 ymax=?
g = - 9.8
m/s/s
2
2
1
2
(15.97)(1.63) 4.9(1.63)
oyy v t gt
y
y
= +
= −
= 13.01 m

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Projectile motion by umakant bhaskar gohatre

  • 1. MR. UMAKANT BHASKARRAO GOHATRE Projectile Motion
  • 2. What is projectile? Projectile -Any object which projected by some means and continues to move due to its own inertia (mass).
  • 3. Projectiles move in TWO dimensions Since a projectile moves in 2- dimensions, it therefore has 2 components just like a resultant vector. Horizontal and Vertical
  • 4. Horizontal “Velocity” Component NEVER changes, covers equal displacements in equal time periods. This means the initial horizontal velocity equals the final horizontal velocity In other words, the horizontal velocity is CONSTANT. BUT WHY? Gravity DOES NOT work horizontally to increase or decrease the velocity.
  • 5. Vertical “Velocity” Component Changes (due to gravity), does NOT cover equal displacements in equal time periods. Both the MAGNITUDE and DIRECTION change. As the projectile moves up the MAGNITUDE DECREASES and its direction is UPWARD. As it moves down the MAGNITUDE INCREASES and the direction is DOWNWARD.
  • 6. Combining the Components Together, these components produce what is called a trajectory or path. This path is parabolic in nature. Component Magnitude Direction Horizontal Constant Constant Vertical Changes Changes
  • 7. Horizontally Launched Projectiles Projectiles which have NO upward trajectory and NO initial VERTICAL velocity. 0 /oyv m s= constantox xv v= =
  • 8. Horizontally Launched Projectiles To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2. 21 2oxx v t at= + oxx v t= Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO! 21 2 y gt= Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.
  • 9. Horizontally Launched Projectiles Example: 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? What do I know? What I want to know? vox=100 m/s t = ? y = 500 m x = ? voy= 0 m/s g = -9.8 m/s/s 2 2 2 1 1500 ( 9.8) 2 2 102.04 y gt t t t = → − = − = → = 10.1 seconds (100)(10.1)oxx v t= = = 1010 m
  • 10. Vertically Launched Projectiles Component Magnitude Direction Horizontal Constant Constant Vertical Decreases up, 0 @ top, Increases down Changes Horizontal Velocity is constant Vertical Velocity decreases on the way upward Vertical Velocity increases on the way down, NO Vertical Velocity at the top of the trajectory.
  • 11. Vertically Launched Projectiles Since the projectile was launched at a angle, the velocity MUST be broken into components!!! cos sin ox o oy o v v v v θ θ = =vo vox voy θ
  • 12. Vertically Launched Projectiles There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0
  • 13. Vertically Launched Projectiles You will still use kinematic #2, but YOU MUST use COMPONENTS in the equation. cos sin ox o oy o v v v v θ θ = = vo vox voy θ oxx v t= 21 2oyy v t gt= +
  • 14. Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? (b) How far away does it land? (c) How high does it travel? cos 20cos53 12.04 / sin 20sin53 15.97 / ox o ox oy o oy v v v m s v v v m s θ θ = = = = = = v o =20.0 m /s θ = 53
  • 15. Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? What I know What I want to know vox=12.04 m/s t = ? voy=15.97 m/s x = ? y = 0 ymax=? g = - 9.8 m/s/s 2 2 2 1 0 (15.97) 4.9 2 15.97 4.9 15.97 4.9 oyy v t gt t t t t t t = + → = − − = − → = = 3.26 s
  • 16. Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (b) How far away does it land? What I know What I want to know vox=12.04 m/s t = 3.26 s voy=15.97 m/s x = ? y = 0 ymax=? g = - 9.8 m/s/s (12.04)(3.26)oxx v t= → = 39.24 m
  • 17. Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (c) How high does it travel? CUT YOUR TIME IN HALF! What I know What I want to know vox=12.04 m/s t = 3.26 s voy=15.97 m/s x = 39.24 m y = 0 ymax=? g = - 9.8 m/s/s 2 2 1 2 (15.97)(1.63) 4.9(1.63) oyy v t gt y y = + = − = 13.01 m