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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 (-)
        ©2011 Creative Curriculum for Children [CCC]
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

      ©2011 Creative Curriculum for Children [CCC]
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.




 ©2011 Creative Curriculum for Children [CCC]
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.”




  ©2011 Creative Curriculum for Children [CCC]
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.




         ©2011 Creative Curriculum for Children [CCC]
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! =]
          ©2011 Creative Curriculum for Children [CCC]
Stopping point 




             Starting point 




     Point after 4 seconds 




©2011 Creative Curriculum for Children [CCC]
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.




        ©2011 Creative Curriculum for Children [CCC]
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:



          ©2011 Creative Curriculum for Children [CCC]
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).


       ©2011 Creative Curriculum for Children [CCC]
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

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Introduction to Kinematics

  • 1. 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
  • 2. 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]
  • 3. 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 (-) ©2011 Creative Curriculum for Children [CCC]
  • 4. 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 ©2011 Creative Curriculum for Children [CCC]
  • 5. 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. ©2011 Creative Curriculum for Children [CCC]
  • 6. 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.” ©2011 Creative Curriculum for Children [CCC]
  • 7. 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. ©2011 Creative Curriculum for Children [CCC]
  • 8. 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! =] ©2011 Creative Curriculum for Children [CCC]
  • 9. Stopping point  Starting point  Point after 4 seconds  ©2011 Creative Curriculum for Children [CCC]
  • 10. 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. ©2011 Creative Curriculum for Children [CCC]
  • 11. 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: ©2011 Creative Curriculum for Children [CCC]
  • 12. 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). ©2011 Creative Curriculum for Children [CCC]
  • 13. 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