Professor Dave again, let's practice
solving kinematics problems.
We've looked at the motion of objects in one
and two dimensions, and hopefully the
examples we discussed made perfect sense.
But many students find that they
understand the examples when they are
explained but have trouble figuring out
what to do on their own during homework
or on a test. The solution to this is to
simply practice a lot and use critical
thinking skills. We will look at a
handful of additional examples now and
as we go through these the key will
always be to ask the following questions:
Number one: what data do I have and what
variables are used to represent this
data? Number two: what is it that I am
trying to find out? Three: what equation can I use
that will allow me to plug in known
values for every variable except the one I
want to solve for? Notice that each of
these three equations is missing one of
the common variables. This one is missing
position
this one is missing final velocity and
this one is missing time.
This allows us to do calculations even
when there is an unknown parameter. And
four: once you arrive at your answer
does it make sense logically or is it
way too big or too small to be
reasonable? If we are able to think
critically in this manner we will find
that many questions in kinematics become
relatively trivial to answer. Let's give
one a shot. Say you toss a ball directly
upward with an initial velocity of 5 m/s.
What will be the maximum height of the
ball? Now as we discussed let's first
decide what we know. We know the
acceleration due to gravity, we know the
initial velocity, here's the clever part
we also know the final velocity because
at the peak of the trajectory the
velocity will be 0. We are solving for
the position associated with the ball
when it will be at its highest point and
at that point the velocity is precisely
zero before it then starts moving in the
negative direction. So which
equation has all the things we know plus
the thing we want to solve for? That
would be this equation where final
velocity squared equals initial velocity
squared plus 2ax. This is convenient
because time is not present, since we
have no value for time. Let's plug in the
numbers,
solve for x and we should get about 1.28
meters as the maximum height of the
ball with respect to the point of
release. If we wanted to know its height
above the ground we would just add
whatever distance there was between the
point of release and the ground. Now
let's say we want to know it's hang time
or how long it is in the air from the
moment you throw it to the moment you
catch it at the same point of release.
To do this we can use the following
equation and solve for the time elapsed
to reach the maximum height and then
simply double it as falling will take
exactly the same amount of time as
ascending. We can use this equation
because we know both the initial and
final velocities and the acceleration
plugging in those values and solving for t
gives us 0.51 seconds. That's how long
it takes to reach the maximum height.
SImply double that and we get 1.02
seconds as the total hang time.
Let's try one more. Say a helicopter is
delivering a care package to a remote
location.
The helicopter is traveling horizontally
at 67 meters per second and is at a
height of 120 meters
when the package is dropped.
How long before the package hits the
ground and how far will it travel in the
horizontal direction? So the first thing
we need to recall is that the vertical
and horizontal motion are independent of
one another.
The package will take the same amount of
time to hit the ground whether there is
horizontal motion or not so we can
isolate the Y direction to find the time
elapsed in the air. We know that the
position will be negative 120 meters
since it will travel that far in the
negative direction, initial velocity in
the y-direction is zero since the
helicopter has no vertical motion, and
acceleration is known.
Let's plug these into this equation and
solve for t which will be about 4.9 five seconds.
Now that we know how long the package is
in the air it is a trivial matter to
calculate how far it will travel in the
horizontal direction. We simply use the
definition of displacement. It is
traveling with the same velocity as the
helicopter the entire time it is in the
air since there is no acceleration
operating in the horizontal direction so
67 meters per second times 4.95 seconds
equals about 332 meters.
So remember that when solving problems
of this nature always ask yourself what
information you have. Also remember that
sometimes you may have information that
is not explicitly listed in the question.
If something is moving from rest the
initial velocity is zero. At the moment
that an object changes direction with
respect to its vertical motion the
velocity is also 0.
Make sure you think critically in this
regard.
Then ask yourself what the question is
looking for and select an equation that
will allow you to solve for this unknown
value. Let's check comprehension.
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