Projectile Motion & Vectors
Test Review
TAKE GOOD NOTES!!
YOU MAY USE ONLY THESE
NOTES ON THE TEST
TOMORROW!!

Projectile motion is the motion of an
object in flight including the impact of
gravity

The path taken in flight is known as a
parabola
Marble/Ramp Vectors
Marble travels in
the x direction
Marble rolls down
the ramp
Marble travels in
the x direction
Vectors

A vector is a drawing showing direction
and magnitude
45 mph east
9.8 m/s2
down
A Thrown ball

The quarterback throws the ball at 9 m/s at a
30° angle (to horizontal)
A Thrown ball


The ball moves in the x direction
The ball moves in the y direction
Vectors can add together


Vectors can work together to describe
the final, or resultant, vector
For example


A boat traveling down a river gets to go
faster because the river “pushes” the boat
faster
A boat traveling up river goes slower
because it has to go against the river
River Trip
River travels
10 mph
downstream
Boat
travels
35 mph
up river
Result –
boat
travels
25 mph
up river
River Trip
River travels
10 mph
downstream
Boat
travels
35 mph
down
river
Result – boat
travels 45
mph down
river
Vectors can add in any
direction!
River travels
10 mph
downstream
River travels
10 mph
downstream
Boat
travels
35 mph
across
the river
Resultant
Boat
travels
35 mph
across
the river
Resultant
You Practice – draw vectors
and calculate resultant



An airplane flies at 255 mph with a 45
mph tailwind (from behind)
A canoeist paddles at 15 mph up river,
while the river flows 3 mph the other
way
A swimmer swims at 6 mph across
(perpendicular) a river flowing at 2 mph
– use a2 + b2 = c2 when vectors are at
right angles to each other
You Practice –
Answers

An airplane flies at 255 mph with a 45
mph tailwind (from behind)
255 mph
Resultant = 300 mph
45 mph
You Practice –
Answers

A canoeist paddles at 15 mph up river,
while the river flows 3 mph the other
way
15 mph
Resultant = 12 mph
3 mph
You Practice –
Answers

A swimmer swims at 6 mph across
(perpendicular) a river flowing at 2 mph


a2 + b2 = c2
62 + 22 = c2 or 40 = c2
6 mph
2 mph
Resultant = 6.32 mph
Calculating the vectors

The y vector =
sin of angle X
velocity

sin(30)*9 m/s

Vy = 4.5 m/s
When the football is thrown, it goes
upwards at 4.5 m/s

30°
Calculating the vectors

The x vector =
cos of angle X
velocity

cos(30)*9 m/s

Vx = 7.79 m/s
When the football is thrown, it goes
down field at 7.79 m/s

30°
Vector formulas

Vx = cos θ * original velocity


θ is the angle from horizontal
Vy = sin θ * original velocity
You Practice – vectors

Find the x and y
vectors for the
football thrown
as shown
V = cos(50)*12 m/s
Vx = 7.71 m/s
 x
V = sin(50)*12 m/s
Vy = 9.19 m/s
 y
50°
What if it is launched
horizontally (no Vy)?

Here is a sample flight with no starting Vy
Distance vs. Time of a Horizontal Launched Object
120
Height (m)
100
80
60
Series1
40
20
0
0
1
2
3
4
Time (seconds)

This is the marble lab we did Monday!!
5
Horizontal Launching



A marble rolls off a
table 1.5 m high with
a velocity of 5 m/s
How far from the
table will it hit the
floor?
Formulas


d=1/2at2
v=d/t
5 m/s
1.5 m


Use d=1/2at2
to find time it drops (and flies
away from the table)





1.5 m = ½(9.8 m/s2)(t2)
3m/(9.8 m/s2)= t2
t2 = .31 seconds2
t =.56 seconds
Find the distance


5 m/s=d/.56 seconds
2.8 m = d
5 m/s
1.5 m
You Practice –
Horizontal Launching



A marble rolls off a
table 3.2 m high with
a velocity of 2.5 m/s
How far from the
table will it hit the
floor?
Formulas


d=1/2at2
v=d/t
5 m/s
1.5 m
You Practice - Answers


Use d=1/2at2
to find time it drops (and flies
away from the table)





3.2 m = ½(9.8 m/s2)(t2)
6.4m/(9.8 m/s2)= t2
t2 = .65 seconds2
t =.81 seconds
Find the distance


2.5 m/s=d/.81 seconds
2.0 m = d
5 m/s
1.5 m
Using vectors


A projectile has
a curved path as
it flies
It spends half of
its flight time on
the way up, and
half on the way
down
50°
Using vectors

Let’s find how
long it takes for
the ball to reach
the top of its
trajectory, or
curved path
V = cos(50)*12 m/s
Vx = 7.71 m/s
 x
V = sin(50)*12 m/s
Vy = 9.19 m/s
 y
50°
How long does it fly?



First, we know it goes
up at 9.19 m/s
Second, we know
velocity at the top of
the trajectory is 0 m/s
Third, we know that
the upwards velocity
decreases at 9.8 m/s2
due to gravity
50°
How long does it fly?

Use the formula







a=(Vf-Vo)/t
-9.8 m/s2=(0 m/s-9.19
m/s)/t
-9.8 m/s2=(-9.19
m/s)/t
t=(-9.19 m/s)/(-9.8
m/s2)
t=.94 seconds to fly
up
ttotal = 2*.94
ttotal = 1.88 seconds
50°
How far it fly?



First, we now know
the time it flies (1.88
seconds)
Second, we know the
horizontal velocity (vx)
= 7.71 m/s
Use the formula v=d/t



7.71 m/s = d/1.88
seconds
d=7.71 m/s * 1.88 s
d = 14.49 m
50°
You Practice –
Using vectors
25°
V = cos(25)*20 m/s
Vx = 18.13 m/s
 x
V = sin(25)*20 m/s
Vy = 8.45 m/s
 y
You Practice –
How long does it fly?

Use the formula







a=(Vf-Vo)/t
-9.8 m/s2=(0 m/s-8.45
m/s)/t
-9.8 m/s2=(-8.45
m/s)/t
t=(-8.45 m/s)/(-9.8
m/s2)
t=.86 seconds to fly
up
ttotal = 2*.86
ttotal = 1.72 seconds
50°
You Practice –
How far it fly?



First, we now know
the time it flies (1.72
seconds)
Second, we know the
horizontal velocity (vx)
= 18.13 m/s
Use the formula v=d/t



18.13 m/s = d/1.72
seconds
d=18.13 m/s * 1.72 s
d = 31.18 m
50°