In
this chapter, we will study the motion in two
dimensions further. First, we will analyze the path
that any object takes when it is thrown in the air.
We will see that all falling objects follows the same
path. Then, we will analyze the motion that repeats
itself, such as the motion of the propeller.

Objects launched are called projectiles.
The flight of baseballs and basketballs are some
examples of projectile motions. Let's analyze the
projectile motion by breaking down the forces acting
on the object.

Suppose a ball was thrown horizontally at the
velocity of 5 m/s.

Let's think about forces acting on x-direction, or
horizontal direction, only. We know that the ball was
thrown at the velocity of 5 m/s. Because no other
force acts on the ball in the air, we know the
velocity will stay constant (because the net force is
zero). Therefore, after 5 seconds, the ball is

d = vt = 5 m/s * 5s = 25 m

25 m away. After 10 seconds,

5 m/s * 10 s = 50 m

The ball is 50 m away.

QUESTION:
How far will the ball travel in 3 seconds in
horizontal direction? m

Let's now think about the forces in
y-direction, or vertical direction. Because gravity
is the only force acting on it, the ball will
accelerate at the rate of -9.8 m/s^{2}, i.e.
9.8 m/s^{2} downward. After 2 seconds, the
ball has traveled 19.6 m because

= 0 + 4.9 * 2^{2}
= 19.6 m

QUESTION:
How far has the ball traveled after 4 seconds
in vertical direction? m

Now, let's combine two forces at x
and y directions and see its displacement.
When t = 0, i.e. right after the ball was thrown, the
displacement on x and y direction are both 0. When t
= 1, after 1 second, the displacement on x direction
is 5 m and on y direction is -4.9 m. When t = 2, the
ball traveled 10 m on x direction and -19.6 m on y
direction. To summarize,

Time
vs. Displacement

Time

Displacement on X

Displacement on Y

0 s

0 m

0 m

1 s

5 m

-9.8 m

2 s

10 m

-19.6 m

QUESTION:
What is the displacement of the ball in x
direction after 3 seconds? m

QUESTION:
What is the displacement of the ball in y
direction after 3 seconds? m

Now, let's look at velocity
rather than displacement. When t = 0, the velocity on
x direction is 5 m/s, while 0 m/s on y direction.
Therefore, the total velocity of the ball is 5 m/s
using Pythagorean Theorem. When t = 1, while the
velocity on x direction stays the same, the velocity
on y direction becomes -4.9 m/s. Therefore, the total
velocity of the ball when t = 1 would be

m/s

To
summarize,

Time vs.
Velocity

Time

Velocity on X

Velocity on Y

Total Velocity

0

5 m/s

0 m/s

5 m/s

1

5 m/s

-4.9 m/s

7.0 m/s

QUESTION:
What is the velocity of the ball in x
direction when t = 2? m/s

QUESTION:
What is the velocity of the ball in y
direction when t = 2? m/s

QUESTION: What
is the total velocity of the ball when t = 2?
m/s

The left drawing
summarizes what we've learned so far. The red
dot represents the positions of a falling
ball with 0.5 seconds of interval.

You can
see that it moves at constant velocity on x
direction, while accelerates on y direction.
As a result, the path of the ball will follow
the path of parabola.

Motion that repeat itself is called periodic
motion. A projectile motion, since it doesn't
repeat, is not periodic. The motion of a swinging
ball is an example of circular motion. Let's analyze
the circular motion.

Consider a planet around the Sun.

When there is no external force, an object will
travel in a straight line (Newton's First Law of
Motion). In order for an object to travel in a
circle, there has to be a force that makes it travel
in a circle.

An object in circular
motion always tries to move in a straight line (Law
of Inertia). However, there is a force that acts toward
the center of the motion. This force is called
the centripetal force. It is the centripetal
force that makes an object travel in a circular path.
This force could be friction or a gravitational
force, but we call it a centripetal force.

When the centripetal force is too strong, the ball
will accelerate toward the center of the circle. When
the centripetal force is too weak, the ball will get
out of the orbit. An object will maintain a circular
motion only when the centripetal force is well
balanced.

These formulas can be used to calculate the
centripetal force:

Where:

F_{c}: is the centripetal force (N)
m: is the mass (kg)
a_{c}: is the centripetal acceleration
(m/s^{2})
v: is the speed (m/s)
r: is the radius of the circle (m)
t: is the period (the length of time required for
one complete rotation) (s)

Too complicated? Let's see a concrete example.

You are rotating with Earth, so you are
experiencing a circular motion.

Say you have a mass of 50 kg. We know the radius
of the circle (6,378,000 m = radius of Earth) and its
period (24 hours = 86400 seconds). From this
information, you can calculate velocity,
acceleration, and centripetal force acting on you.

To calculate velocity,
use the formula .

m/s = 1670 km/h.

To calculate acceleration, use the formula .

m/s^{2}.

You can see that we don't accelerate much (which
is obvious).

QUESTION:
A ball with a mass of 250 g has a centripetal
acceleration of 5 m/s^{2}. What is
the centripetal force acting on this ball? N

QUESTION:
A ball with a speed of 10 m/s is in a
circular motion. If the circle has a radius
of 20 m, what would the period of the ball
be? s

QUESTION:
The Moon's orbit around earth is nearly
circular. The orbit has a radius of about
385,000 km and a period of 27.3 days.
Determine the acceleration of the moon toward
Earth. m/s^{2}