Physics Unit 5
Lesson 1
Projectile Motion
Falling things, and rockets ‘n’ that…
Projectile Motion
If an object is launched at an initial angle of θ0 with the horizontal, the
analysis is similar except that the initial velocity has a vertical component.
Initial Vertical Velocity does NOT = ZERO
Initial Horizontal Velocity does NOT = Zero
t = 2 Vy /g (if Y = 0)
Projectile Motion
It can be understood by analyzing
the horizontal and vertical motions
separately.
Initial Vertical Velocity = ZERO
t = √{2y/g} (if Vy =0)
Initial Horizontal Velocity = Zero
Projectile Motion Equations
Horizontal Component:
X = Vx(t)
Vx = Vi COS 
Vertical Component:
Y = Vy(t) –(1/2)gt2
Vy = Vi SIN 
t = 2 Vy /g (if Y = 0)
t = √{2y/g} (if Vy =0)

Homework– Glencoe Page 150151 #’s 1 - 5 ALL
Objectives:
Unit 5
Lesson 2:
Do Now:
How far above the ground will a ball be
above the ground, if it is thrown from
ground level @ 30.625 m/s straight up in
Y = Vi (t) – ½ gt2
2.50 seconds? Use 9.81 m/s2 for gravity Y = 30.625 (2.5) – ½ (9.81)(2.5)2
Y = 76.5625 – 30.625
Y = 45.9375 m = 45.9 m
:
Review - Projectile Motion
Equations
Horizontal Component:
X = Vx(t)
Vx = Vi COS 
Vertical Component:
Y = Vy(t) –(1/2)gt2
Vy = Vi SIN 
t = 2 Vy /g (if Y = 0)
t = √{2y/g} (if Vy =0)
Projectile Motion
• Projectile motion follows a parabolic path i.e.
http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html
• It has a horizontal (x) component and a vertical (y)
component.
• The horizontal component is not affected by gravity,
so use separate equations for the different
components.
Continued…
• What does this mean for objects dropped from the
same height?
•
Objects dropped from the same height will hit the ground at
the same time, regardless of weight (ignoring differences in air
resistance)
http://www.animations.physics.unsw.edu.au/jw/projectiles.htm#1
• Solve the same way you would any other kinematics
problem… AICR
• State what you know, what do you want to find?
• Choose an equation and solve for your unknown.
• Look out for key words like..
“dropped”, “from rest” and “stationary:”
Solving Problems Involving
AICR
Projectile Motion
1. Read the problem carefully, and choose the object(s) you are going to analyze.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same in both directions, and includes
only the time the object is moving with constant acceleration g.
5. Examine the x and y motions separately.
6. List known and unknown quantities. Remember that vx never changes, and that
vy = 0 at the highest point.
7. Plan how you will proceed. Use the appropriate equations; you may have to
combine some of them.
Projectile Motion Lesson 3
Objectives:
Gain proficiency in collecting Data to solve for
2-D motion variables:
Do NOW:
How far down range will a cannonball go if fired
at an angle of 30.0 deg
an initial velocity of 18.5 m/s?
assume no air resistance.
What will hit the
ground first a bullet
dropped or a bullet
fired from the same
height?
Discuss with your neighbor…
http://www.youtube.com/watch?v=D9wQVIEdKh8
Objects Under the Influence
of Gravity:
Projectile Motion
The speed in the x-direction is constant; in
the y-direction the object moves with
constant acceleration g.
This photograph shows two balls
that start to fall at the same time.
The one on the right has an initial
speed in the x-direction. It can be
seen that vertical positions of the
two balls are identical at identical
times, while the horizontal
position of the yellow ball
increases linearly.
Projectile Motion Day Two
Problems
Projectile Motion Water slide
Water Slide 2
Giant House Slide
DRIVING OFF A CLIFF!!
How fast must the motorcycle leave the cliff to land at
x = 90 m, y = -50 m? vx0 = ?
y is positive upward
y0 = 0 at top
vy0 = 0
vx = vx0 = ?
x = vx0t
vy = -gt
y = - (½)gt2
Time to Bottom:
t = √2y/(-g) = 3.19 s
vx0 = (x/t) = 28.2 m/s
A PUNT!
FIND HANG TIME:
RANGE:
MAXIMUM HEIGHT:
Hang Time:2.53 seconds
Range: 40.5 m
Max Height: 8.39 m
v0 = 20.0 m/s, θ0 = 37.0º
vx0= v0cos(θ0) = 16.0 15.97 m/s
vy0= v0sin(θ0) = 12.036 m/s
y = y0+ vy0 t – (½)g t2 t= 2.531 sec (quadratic formula)
x = vx0 t
x = 40.540.49 m
t/2 = 1.27 sec
Ymax = y0+ vy0 t – (½)g t2 = 1+ 12.0 (1.27 ) – 4.9 (1.27 )2
Y max = 8.3926 m
Test Review Unit 5
Motion in One-Dimension
Displacement / Instantaneous Velocity / Acceleration
Motion Diagrams
Free Falling Objects / 1-Dimensional with constant Acceleration
Motion in Two-Dimensions
Vectors Coordinate System / Properties of Vectors
Components of Vectors / Unit Vectors
Adding Subtracting Multiplying Vectors (Graphically and Analytically)
Scalar Products - Dot Products – (Honors: Cross Products)
Motion in Two-Dimensions
Projectile
Range / Velocity Components / Maximum Height
Time of Flight
Circular Motion