Introduction to Kinematics

Creative Curriculum for
Children: PHYSICS
Kinematics

Projectiles here, there, there &
everywhere.

Product of Victor Hakim, C.C.C. Vice President and Director of Physical Sciences

One of the most common types of problems in physics
is a branch called kinematics. This deals with
trajectories and the details of how an object acts when
in motion.

In Greek, kinema- means “movement” and that’s
exactly what we’re studying:

<<
What’s the maximum height after throwing an object in
the air? What’s the maximum velocity? What will be the
velocity after n seconds?
>> We’ll learn how to answer these today!

©2011 Creative Curriculum for Children [CCC]

First, we need a few pieces of key information:

We must measure everything in what are called S. I. Units
(International System of Units in French). Time is measured in
seconds and displacement in meters. These yield velocity units of
m/s and acceleration units in m/s/s or m/s2.

We must set up our coordinate axes so that the y-axis is going up
and down, and the x-axis is going left and right. Up and right are
positive (+), and down and left are negative (-).
y (+)

x (-) x (+)

y (-)

Because of Earth’s gravity, there will always be an acceleration
of objects downward, and we use this in our calculations as a
negative number. Objects will decelerate, or lose an amount of
velocity after every second, and it is always the same loss of
velocity after each second.

The value of Earth’s gravitational pull is always constant (close
to the surface of the Earth) and we use equations for kinematics
only on the assumption that the acceleration downwards never
changes. We happen to call this constant downwards
acceleration g, a value equal to about 9.81 m/s2. But we always
use g = 10 m/s2. With this substitution, calculations become
unbelievably easier. Because the acceleration points
downward, the value of acceleration
for projectiles is -g.

Acceleration = -10 m/s2


Now for the equations.
First, it’s very important that you understand the
meaning of each variable:

• t is the variable that stands for the time. It is
measured in seconds.
• v stands for velocity. Measured in m/s, this can
be either positive or negative.
• g stands for the acceleration on Earth. This is
measured in m/s2. For problems on the Earth’s
surface, the value will always be g = 10 m/s2.


Don’t worry about memorizing these formulas just
yet, but know how to use them and become familiar with
their applications.

1. v = v0 – gt
v0 2
2. Maximum Height = 2g
+ H0

These equations consider the velocity in the y-axis
direction. In other words, they only work for velocities
that are straight up or straight down.

Equation (1) tells us the velocity, v, at any time t in
seconds. v0 is a constant. It stands for the initial velocity
with which a projectile is thrown, namely, at time t=0. We
say, “V initial.”


Equation (2) tells us the maximum height a projectile will go
given an initial vertical velocity (+ only). Also, H0 refers to the
initial height, if I’m not shooting the projectile initially (at time
t=0) from the ground. I can shoot from atop Mount
Everest, 8,848 meters high, but I can still find my maximum
height by setting H0 = 8,848.

Equation (2): I’ll throw the ball up at
initial velocity v0, and I want to find
the maximum height that that ball will
go.


Let’s try some examples.
I’ll throw a projectile up at 20 m/s. What is its velocity after 4
seconds?

Use Equation (1). My v0 = 20. g, as always, will be equal to 10.
Now, if I’m asking for the velocity after 4 seconds, simply set t=4
and solve for v:

v = 20 – 10x4
= -20 m/s2.

Does this make sense? I’ll throw the ball upwards, because v0
was positive, and the Earth will slow it down until it stops.
Then, the ball will come down, and so it should be at a negative
velocity.

We got a negative velocity! =]

Stopping point 

Starting point 

Point after 4 seconds 


Maximum Height.
I am 100 meters tall. I shoot the ball up, from my head, at 5 m/s.
Find the maximum height.

I think you get the idea. H0 = 100 and v0 = 5. Plug them in into
Equation (2):

Max height = 52/(20) +100 You can use a calculator!
= 101.25 m.

Notice that the max height is only 1.25 m above the starting
height. That makes sense because I only threw it at 5 m/s, a
velocity that’s not too fast.


A few more tips…
• It’s important to understand the problem. Picture the
situation, create it, and this will help you to both set up the
problem and verify that you’re final answer makes logical
sense.

• Become familiar with the equations so that when you get a
problem, you can tell right away which equation that’s good
for. Physics is all about knowing which formulas to use. Pick
the right one with the variables that are given to you.

• Next time, we’ll experiment with projectiles shot not perfectly
vertically, but also angled a little to the side:


Practice—good luck!
1. If I shoot a ball up from the ground at 82 m/s, after how
many seconds will it temporarily stop moving?

2. From problem 1, what is the max height?

3. I shot a projectile, and it reached its max height of 20
meters. With what velocity did I initially throw it up?

4. If it takes 10 seconds for a ball to drop to the ground
from atop a building, tell me the velocity with which the
ball hit the ground (this will be the maximum magnitude
of velocity in the entire experiment).


Answers #1-4

1. 8.2 seconds
2. 336.2 meters
3. 20 m/s
4. -100 m/s

Product of Victor Hakim, C.C.C. Vice President and Director of Physical Sciences

Introduction to Kinematics

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