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…
A shortcut: 
For AP multiple choice, since you may not
use a calculator, almost always g = 10 ms2.
So for anything launched horizontally,
Vertical: d = vot + ½ at2 with vo = 0
Height, H = ½ gt2 = 5t2
But… at the highest point of a projectile
launched at an angle, it is moving with
ONLY a horizontal velocity… so… H = 5t2
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