Projectile Motion
Physics
Projectiles!
Projectile – any air born object with only
the force of gravity acting on it.
Examples – basketball thrown in air, baseball hit with
bat in air, speeding bullet
Non-examples – basketball while still in hands,
baseball rolling on the ground
Characteristics of Projectiles:
Its path is called its trajectory
Follow special paths called parabolas

Remember, from math class?
Ignore air resistance

Think Frictionless (no air resistance)
Characteristics of Projectiles:
Vertical velocity is changing with time

a = g = 9.8 m/s2 downward
Projectile Motion


Horizontal velocity is independent of vertical
Horizontal velocity is ALWAYS constant
(Vi,x = vf,x so no acceleration)
Factors Affecting Projectile Motion
What two factors would affect projectile
motion?


Angle
Initial velocity
Initial
Velocity
Angle
Problem Solving Steps
1. Select coordinate system

Typically initial position of projectile is (0,0)
2. Examine Horizontal (x) and Vertical (y) Motion
- Separate initial velocity into x & y components
v
vy
vx
Problem Solving Technique
3. X and Y motions are connected by the common
time.
- only consider time the object is in air
4. Treat horizontal and vertical motion separately
- List unknown and known quantities in chart
- apply equations
- cannot mix variables …. Variables from the horizontal
motion cannot be used in the same equation with
vertical variables
Revised FORMULAS !!!!
Have a formula sheet?
Make chart with two columns
Horizontal
Vertical
Horizontal Motion
dx = vx t
position (horizontal) = Velocity (horizontal) * time
vx = vx,i = vx,f
all are constant!
Velocity(horizontal) = constant
Velocity(horizontal,initial) = Velocity(horizontal,final)
Think frictionless (no air resistance)
VERTICAL MOTION
vf,y = vi,y + at
Velocity(final,vertical) = velocity(initial,vertical) + acceleration*time
vf2,y = vi2,y +2ady
Velocity(vertical,final squared) = Velocity(vertical,initial squared) + 2*
acceleration*displacement(vertical)
dy = vi,yt + ½ at2
Displacement(vertical) = Velocity(initial,vertical)* time+
(1/2)*acceleration* time squared
acceleration(a) = gravity(g) = 9.8 m/s2 downward
Practice Problem 1
A plane is 10.0m above the ground, traveling at a velocity of 22.5
m/s in the positive x direction. The pilot wants to know where in
the plane’s path should the stunt dummy be dropped so that it
will land in the swimming pool.
1) Make a chart with 2 columns: Horizontal and Vertical
2) List of givens and unknowns
Horizontal Vertical
3) Solve for unknowns
x
d
y
?
vi 22.5 m/s
vf
10 m
0m
22.5 m/s
a 0 m/s2
t
9.8 m/s2
Practice Problem 1
A plane is 10.0m above the ground, traveling at a velocity of 22.5 m/s in
the positive x direction. The pilot wants to know where in the plane’s
path should the stunt dummy be dropped so that it will land in the
swimming pool.
3) Solve for unknowns
Horizontal Vertical
x
y
d ? 32.14m 10m
Vi 22.5 m/s
0m
Vf 22.5 m/s
A
t
0 m/s2
1.43 s
-9.8 m/s2
1.43 s
Would like to do v = d/t but don’t know
time. 1st step is to get time
1st step …
dy = ½ at2+vit
10.0 = ½(-9.8)t2
10/-4.9 = t2
1.42857… s = t
2nd step …
dx = vxt
dX = (22.5)(1.43)
dX = 32.14m
Practice Problem 2
An object is fired from the ground at 100 m/s
at an angle of 30º with the horizontal.
a. Calculate the horizontal and vertical components of
the initial velocity.
b. After 2.0 seconds, how far has the object traveled in
the horizontal direction?
c. How high is the object at this point (at 2.0s) ?
Practice Problem 2
Part a
 Draw diagram
 Sin(30) = y/100 = 50 m/s in y or vertical direction
 Cos(30) = x/100 = 86.6 m/s in x or horizontal direction
Part b
 dx = vxt
 dx = (86.6 m/s)(2.0s) = 173 m
Part c
 dy = ½at2 + vyt
 dy = ½(-9.8)(2)2 + (50 m/s)(2.0s)
 dy = -19.6 +100 = 80.4 m