Motion in Two Dimensions
LAB FOLDERS DUE
TOMORROW!
(Ignore any effects from air resistance)
A pickup is moving with a constant velocity
and a hunter is sitting in the back pointing
his rifle straight up.
If he fires the rifle,
where will the bullet land?
In front of the pickup?
Behind the pickup?
Inside the pickup?
A rifle, at a height H above the ground, fires a
bullet parallel to the ground.
At the same instant and from the same height H,
a second bullet is dropped from rest.
Which bullet strikes the ground first?
While riding in a car traveling down the road, you
reach out and drop a cat. What would its
pathway to the ground look like to you?
What would the cat’s pathway look like to a
person standing alongside the road?
Projectiles
Projectile: an object that moves through air or
space, acted upon only by gravity
Range: the horizontal distance traveled by a
projectile
Trajectory: the pathway of a projectile
Gravity provides acceleration in the vertical
direction.
But…..
Assuming no air resistance, there is NOTHING
to affect its horizontal velocity so…
The horizontal acceleration is ZERO
Drop an object or launch it horizontally?
Which one hits the ground first?
Gravity acts the same on both!
An object dropped from a given height
will hit the ground
AT THE SAME TIME
As an object launched horizontally from
the same height.
Why?
Gravity pulls both of them to the ground
at the same rate.
Acceleration due to gravity = -9.8 m/s/s
Projectiles Launched Horizontally
d = vot + ½ at2
V = 20 m/s
x (horizontal )
vo = 20 m/s
vo = 0
a = - g m/s2
a=0
Range (horizontal distance) = dx = vot
y (vertical)
dy = Height = ½ gt2
H = ½ (9.8)t2
H = ½ (9.8)t2
H = 50 m
t 
Range (horizontal distance) = dx
Range = 20 m/s x 3.19 s = 63.8 m
H

1
g
2
50
 3.19 s
4.9
Projectiles Launched Horizontally
V = 30 m/s
d = vot + ½ at2
Horizontal distance = dx = vot
H = 50 m
Range (horizontal distance) = dx
Range = 30 m/s x 3.19 s = 95.7m
Height = ½(9.8)t2
Projectiles Launched Horizontally
V = 10 m/s
H = 50 m
Range (horizontal distance) = dx = vt
Range = 10 m/s x 3.19 s = 31.9m
Projectiles Launched Horizontally
V = ???
H = ½ gt2
t
H
1
g
2
Horizontal distance = dx = vt
dx
v
t
Projectiles Launched Horizontally
Time to ground = 3.19 seconds!
H = 50 m
Think vertical
Think horizontal
A cannon ball is shot horizontally at 45 m/s
off the top of a cliff 100 m high. How long
did it take for the ball to hit the ground?
How far did it land from the base of the
cliff
Height = ½ gt2
horizontal d = vot
H ÷ ½(9.8) = t2
t = 4.52 s
d = vot
d = 45 m/s (4.52 s)
= 203.4 m
Think vertical
Think horizontal
• A ball rolling at 8 m/s rolls off the top of a
20 m high building. How long will it take to
hit the ground? How far will it land from the
base of the building?
Height = ½ gt2
horizontal d = vot
H ÷ ½(9.8) = t2
t = 2.02 s
d = v ot
d = 8 m/s (2.02 s) =
16.16 m
Think vertical
Think horizontal
• A projectile hits the ground 3 seconds
after it was launched horizontally from the
top of a hill. If the range of the projectile
was 60 m, how fast was the projectile
launched? How tall was the hill?
Height = ½ gt2
horizontal d = vot
H = ½(9.8) (3)2 = 44.1 m
d = v ot
vo = d ÷ t = 60 ÷ 3
= 20 m/s
Think vertical
Think horizontal
• If a projectile hit the ground 4 seconds
after it was launched and the range was 50
m, how fast was the launch velocity? How
tall was the hill?
Height = ½ gt2
horizontal d = vot
H = ½ (9.8) (4)2 = 78.4 m
d = v ot
vo = d ÷ t = 50 ÷ 4
= 12.5 m/s
Think vertical
Think horizontal
• A bullet is shot horizontally from the top of a
hill. It strikes the ground in 1.5 seconds
with a range of 700 m. How tall was the
hill? How fast was the original velocity of
the bullet?
Height = ½ gt2
horizontal d = vot
H = ½(9.8) (1.5)2 = 11.025 m
d = v ot
vo = d ÷ t = 700 ÷ 1.5 =
466 m/s
LAB BOOKS DUE NOW !!
Projectiles Launched at an Angle
The original velocity, vo of the projectile is
not entirely horizontal nor entirely vertical,
but has COMPONENTS that are both
horizontal and vertical.
vy
vx
The vertical component of the velocity, vy is
affected by gravity and will change at a
rate of g = 10 m/s every second.
(“g” is the magnitude of the acceleration due to gravity, but we understand the
acceleration is in the downward direction)
a = -g = -10 m/s2
The horizontal component of the velocity, vx is
NOT affected by gravity and if there is no air
resistance, it remains constant.
vy
vx
Suppose the vertical component of the
velocity, vy , was 50 m/s. What will it be 1
second after the launch? Remember,
a = - g = -10 m/s per second
Vy = 40 m/s
Vy = 50 m/s
Suppose the vertical component of the
velocity, vy , was 50 m/s. What will it be 2
second after the launch?
Vy = 30 m/s
Suppose the vertical component of the
velocity, vy , was 50 m/s. What will it be 3
second after the launch?
Vy = 20 m/s
Suppose the vertical component of the
velocity, vy , was 50 m/s. What will it be 4
second after the launch?
Vy = 10 m/s
Suppose the vertical component of the
velocity, vy , was 50 m/s. What will it be 5
second after the launch?
Vy = 0 m/s
Suppose the vertical component of the
velocity, vy , was 50 m/s. What will it be 6
second after the launch?
Vy = -10 m/s
vx
Suppose the vertical component of the
velocity, vy , was 50 m/s.
How could one find the time it takes until a
projectile reaches its highest point?
Vy = 50 m/s
vx
Suppose the vertical component of the
velocity, vy , was 60 m/s. What will it be 1
second after the launch? 2 s? 3s? 4s?
5s? 6s? 7s?
How could one find the time it takes until a
projectile reaches its highest point?
Vy = 60 m/s
vx
Suppose the vertical component of the
velocity, vy , was 55 m/s. What will it be 1
second after the launch?
How could one find the time it takes until a
projectile reaches its highest point?
Vy = 55 m/s
vx
Suppose the vertical component of the velocity,
vy , was 40 m/s and the horizontal component of the
velocity, vx, was 30 m/s.
How long will it be until it reaches its highest point?
4s
How much longer until it hits the ground?
4s
How far in the horizontal x direction has the projectile
traveled during those 8 seconds? In other words, what
is its RANGE?
Range = dx = vxt
= (30 m/s) (8 s)
Range = 240 m
Vy = 40 m/s
Vx = 30 m/s
240 m
Suppose the vertical component of the velocity,
vy , was 60 m/s and the horizontal component of the
velocity, vx, was 15 m/s.
How long will it be until it reaches its highest point?
6s
How much longer until it hits the ground?
6s
How far in the horizontal x direction has the projectile
traveled during those 12 seconds? In other words, what
is its RANGE?
dx = vxt
= (15 m/s) (12 s)
Range = 180 m
Vy = 60 m/s
Vx = 15 m/s
180 m
Suppose the vertical component of the velocity,
vy , was 30 m/s and the horizontal component of the
velocity, vx, was 50 m/s.
How long will it be until it reaches its highest point?
3s
How much longer until it hits the ground?
3s
How far in the horizontal x direction has the projectile
traveled during those 6 seconds? In other words, what
is its RANGE?
dx = vxt
= (50 m/s) (6 s)
Range = 300 m
Vy = 30 m/s
Vx = 50 m/s
300 m
Suppose the vertical component of the velocity,
vy , was 45 m/s and the horizontal component of the
velocity, vx, was 30 m/s.
How long will it be until it reaches its highest point?
4.5s
How much longer until it hits the ground?
4.5s
How far in the horizontal x direction has the projectile
traveled during those 9 seconds? In other words, what
is its RANGE?
dx = vxt
= (30 m/s) (9 s)
Range = 270 m
Vy = 45 m/s
Vx = 30 m/s
270 m
Suppose the vertical component of the velocity,
vy , was 45 m/s and the horizontal component of the
velocity, vx, was 30 m/s.
How fast was vo? It’s not 45 m/s. It’s not 30 m/s. How would
you find the original velocity, vo?
Use the Pythagorean Theorem!
452  302  54m / s
Vy = 45 m/s
Vx = 30 m/s
At what part of the trajectory does a
projectile have minimum speed?
At what angle will a projectile go the
farthest?
45 degrees
A ball is launched at an angle of 30 degrees
and hits a target. At what other angle
could the ball be launched and still hit the
target?
45 – 30 = 15 degrees
45 + 15 = 60 degrees
If you throw a ball STRAIGHT up, what is its
velocity at its highest point?
ZERO
What is its acceleration at its highest point?
- 10 m/s2
If you throw a ball at an angle, what is its
velocity at its highest point?
Its velocity at its highest point is NOT zero,
but is the same as its original x component
of its velocity.
What is its acceleration at its highest point?
-10 m/s2
The following 2 slides are for Pre-AP only!
Analyzing projectiles using
vector components
Most of the time the horizontal and vertical
components of vo are not provided and must
be first identified by resolving the initial
velocity vector.
After identifying the components, the
projectile can be analyzed in the same way
for time to the highest point, range, etc.
*Note: Leave the velocity vector components in the
form of vocos q or vo sin q in your calculator
rather than calculating out the value! This keeps
errors to a minimum.
vy = vo sin q
q = 35 degrees
= 12 sin 35˚
vx = vo cos q = 12 cos 35˚
Time to the highest point?
𝑣𝑜 sin 𝜃
12𝑠𝑖𝑛35˚
𝑡=
=
𝑔
Time to ground?
= 0.7 𝑠
9.8
0.7 s x 2 = 1.4 s
Range? 𝑑 = (𝑣𝑜 𝑐𝑜𝑠𝜃)t = (12cos35˚)1.4 = 13.76 m
Max height?
Since vf = 0 (in vertical direction) at the highest position, you can use
vf2 = vo2 + 2ad to find max vertical height or…
Two- Dimension Motion:
Circular Motion
Question: Is there a constant velocity when an
object moves in a circle with a constant speed?
No, the direction changes, therefore the velocity
changes.
If the velocity changed, the object is actually
ACCELERATING even while moving at the same
speed.
The change in direction is always pointed toward the
center of the circle, so the acceleration is inwardtoward the CENTER.
“centripetal”- center seeking
acceleration
Centripetal Acceleration
acceleration
EVERYTHING that moves in a circular path is
experiencing a centripetal acceleration that
points inward- toward the center of the circle!!
a