# Projectile Motion Projectile Motion Projectile motion a combination of

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```					         Projectile Motion
Projectile motion: a combination of horizontal
motion with constant horizontal velocity and
vertical motion with a constant downward
acceleration due to gravity.
Projectile motion refers to the motion of an
object that is thrown, or projected, into the air
at an angle. We restrict ourselves to objects
thrown near the Earth’s surface as the distance
traveled and the maximum height above the
Earth are small compared to the Earth’s radius
so that gravity can be considered to be
constant.
Projectile Motion
The motion of a projectile is determined
only by the object’s initial velocity and
gravity.
The vertical motion of a projected object
is independent of its horizontal motion.
The vertical motion of a projectile is
nothing more than free fall.
The one common variable between the
horizontal and vertical motions is time.
Path of a Projectile
A projectile moves horizontally with
constant velocity while being accelerated
vertically. A right angle exists between
the direction of the horizontal and
vertical motion; the resultant motion in
these two dimensions is a curved path.
   The path of a projectile is called its
trajectory.
   The trajectory of a projectile in free fall is a
parabola.
Path of a Projectile
Path of a Projectile
vo = initial velocity or resultant velocity
vx = horizontal velocity
vyi = initial vertical velocity
vyf = final vertical velocity
R= maximum horizontal distance (range)
x = horizontal distance
Dy = change in vertical position
yi = initial vertical position
yf = final vertical position
q = angle of projection (launch angle)
H = maximum height
ag = gravity = 9.8 m/s2
Path of a Projectile
Path of a Projectile
The horizontal distance traveled by a
projectile is determined by the horizontal
velocity and the time the projectile
remains in the air. The time the
projectile remains in the air is dependent
upon gravity.
Immediately after release of the
projectile, the force of gravity begins to
accelerate the projectile vertically
towards the Earth’s center of gravity.
Path of a Projectile
The velocity vector vo changes with time in both
magnitude and direction. This change is the
result of acceleration in the negative y direction
(due to gravity). The horizontal component (x
component) of the velocity vo remains constant
over time because there is no acceleration along
the horizontal direction
The vertical component (vy) of the velocity vo is
zero at the peak of the trajectory. However, there
is a horizontal component of velocity, vx, at the
peak of the trajectory.
Path of a Projectile
Path of a Projectile
In the prior diagram, r is the position vector of
the projectile. The position vector has x and y
components and is the hypotenuse of the right
triangle formed when the x and y components
are plotted.
 The velocity vector vot would be the
displacement of the projectile if gravity were
not acting on the projectile.
 The vector 0.5agt is the vertical
2

displacement of the projectile due to the
downward acceleration of gravity.
 Together, this determines the vertical
position for the projectile:
Δy = (vy·t) – (0.5·ag·t2)
Path of a Projectile
Problem Solving: Projectile Motion

Analyze the horizontal motion and the
vertical motion separately. If you are
given the velocity of projection, vo, you
may want to resolve it into its x and y
components.
Think for a minute before jumping into
the equations. Remember that vx
remains constant throughout the
trajectory, and that vy = 0 m/s at the
highest point of any trajectory that
returns downward.
Horizontal velocity component:

vx is constant because there is no
acceleration in the horizontal direction if
air resistance is ignored.

v x  vo  cos q
Vertical velocity component:

At the time of launch:

v y i  v o  sin θ
After the launch:

v y f  v o  sinθ   a g  t          
   If vy positive, direction of vertical motion is
up; if vy negative, direction of vertical motion
is down; if vy = 0, projectile is at highest
point.
Horizontal position component:

x  vx  t
x  v o  cos θ  t
If you launch the projectile
horizontally:
    then vo = vx
   vyi = 0 m/s
   q = 0o
Vertical position component:
Δy  y f  y i
 
y f  y i  v yi  t  0.5  a g  t   2


y f  y i  v o  sin θ  t   0.5  a g  t   2

Relationship Between Vertical
and Horizontal Position:

      x ag
2           
Δy  x  tan θ                       
 2  v 2  cos θ2 
      o             
this equation is only valid for
launch angles in the range
0 < q < 90
Range (total horizontal displacement)

v o  sin 2  θ 
2
R
ag
double angle formula :
sin (2  θ)  2  sin θ  cos θ
v o  2  sin θ  cos θ
2
R
ag
Maximum Height

v o  sin θ
2         2
H

2ag  
When Do The Range & Maximum
Height Equations Work?
Works when Dy = 0.   Does not work when
Dy  0.
Determining vo from vx and vy

If the vertical and horizontal
components of the velocity are
known, then the magnitude and
direction of the resultant velocity
can be determined.
Magnitude:
vo  v x  v y
2      2
Determining vo from vx and vy

Direction: from the horizontal
 vy 
θ  tan  
1
v 
 x
Direction: from the vertical
 vx 
1 
θ  tan         
v 
 y
Range and Angle of Projection
Range and Angle of Projection

The range is a maximum at 45
because sin (2·45) = 1.
For any angle q other than 45, a
point having coordinates (x,0) can
be reached by using either one of
two complimentary angles for q,
such as 15  and 75  or 30  and
50 .
Range and Angle of Projection
The maximum height and time of flight
differ for the two trajectories having the
same coordinates (x, 0).

A launch angle of 90° (straight up) will
result in the maximum height any
projectile can reach.
For Objects Shot Horizontally:
vx constant
x              m
vx        v yi  0
t              s     D

x  vx  t
Dy negative
Dy = -height

Δy  0.5  a g  t   2
For Objects Shot Horizontally:
When hits at bottom:
   Vyf should be negative
   vo = resultant velocity

v y f  a g  t

vo  v x  v y f
2      2
For Objects Shot Horizontally:
q with horizontal:                   v yf
1
q  tan
vx

q with vertical:                    vx
1
q  tan
v yf
For Situations in Which Dy = 0 m
v o 2  sin 2  θ 
R
ag
v o 2  sinθ 2
H
2ag
v y i  v o  sinθ
v x  v o  cosθ
R
vx 
t
       
Δy  v y i  t  0.5  a g  t 2   
For Situations In Which Dy Positive
For Situations In Which Dy Positive
v x  v o  cos q            v yi  v o  sin q
At any point in the flight:
       
Δy  v y i  t  0.5  a g  t 2   
      x2 ag          
Δy  x  tanθ                         
 2  v 2  cosθ 2   
      o               

vyf  vyi  ag  t   
x  vx  t
v x constant

vo  v x 2  v y f2
For Situations In Which Dy Negative
For Situations In Which Dy Negative

At launch: v x  v o  cos q         v yi  v o  sin q
After launch: x  v x  t
       
Δy  v y i  t  0.5  a g  t 2   
      x2 ag           
Δy  x  tanθ                          
 2  v 2  cosθ 2    
      o                
v x constant
When it hits ground:
vyf  vyi  ag  t         
vo  v x  v y f
2             2

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