Professor Dave here, let's discuss
vertical motion.
We have familiarized ourselves with position,
velocity, and acceleration, as well as the
kinematic equations that govern the
relationship between these parameters. We
also saw how we can apply these to
simple horizontal motion, and things will
be essentially the same when looking at
vertical motion.
The only difference is that when
examining strictly vertical motion, the
acceleration will be the acceleration
due to Earth's gravity, which is a
constant, -9.8 meters per second squared.
This value is unique to the earth since
the earth has a particular mass and
radius. This value would be different if
we were on some other moon or planet, but
we humans tend to spend most of our time
here on earth, so we can just associate
acceleration due to gravity with this
value unless otherwise specified. Also
with horizontal motion, it is arbitrary
which direction we regard as positive or
negative, though we often assign the
positive direction to the right by
convention, but with vertical motion the
negative direction will always be down
towards Earth, and objects will always
accelerate towards the Earth, since that
is the main source of gravity we
experience every day. We can examine
objects that are in free fall, which
means they were dropped downwards from a
standstill, or we can look at objects
that have some initial upward velocity.
Either way, acceleration will always be
-9.8 meters per second squared
when on earth. Let's say for example you
drop a rock off of a 100 meter tall cliff.
How long will it take to hit the ground
and how fast will it be traveling when
it lands? To answer this we will need the
equation regarding position, because we
know that the rock must travel 100
meters in the negative direction. We can
plug in -100 here. For the
initial velocity that will be 0 because
when you drop the rock
it is from a complete standstill,
only once you let go does gravity cause
the rock to accelerate towards the
ground, so this whole term can be ignored.
Now we plug in the acceleration due to
gravity and solve for t. Notice that the
negative signs on position and
acceleration values will cancel out to
give a positive value for time. This is
very important, as we have to be careful
with signs as we do arithmetic, and this
will be a common opportunity for error
with many students. And we get around 4.5
seconds for the time elapsed during
free fall. Now if we want to know how fast
it is going when it hits the ground, we
can use this equation to solve for
velocity. Again initial velocity is 0 and
we have -9.8 meters per second
per second, times 4.5 seconds which gives us
-44.1 m/s at the time of impact.
This value is negative because velocity
is a vector and the object is traveling
in the negative direction. Now let's say
that we take another rock and this time
we throw the rock straight up in the
air with an initial velocity of 10 m/s.
This velocity will immediately begin to
decrease because of the negative
acceleration due to Earth's gravity. The
moment it leaves your hand the velocity
will decrease until it becomes 0 and
then grows in the negative direction.
This means it will travel upwards for
some amount of time before falling back
down to the bottom where the other rock
landed. How long will this rock be
traveling before it lands?
Well let's do the same calculation that
we did before, the only difference here
is that instead of an initial velocity
of 0 we have an initial velocity of 10
meters per second in the positive
direction, or upwards, and the usual
acceleration which again will be
downwards and therefore negative. We plug
in -100 meters for the
displacement, remembering that even
though the distance this rock will
travel is greater than the other one,
both rocks start and end in the same
place.
Then we plug in the other numbers. What
we are left with is a little trickier
than before because we have both a t
and t squared term. This means that we
will have to use the quadratic equation.
If you have forgotten this from algebra
class, here it is as it applies to an
equation in ax^2 + bx + c
form, we simply take the coefficients of
the equation and plug them into this
expression and we will get two answers
as possible values for x. So we have to
take our equation and rearrange it to
look like the standard form which
involves adding 100 to both sides.
There's our a, b, and c. If we plug those
in we get the two values for x but the
positive answer is the one we are
looking for, as it doesn't make sense to
discuss negative quantities of time, so
we can conclude that it takes 5.65
seconds for this rock to hit
the ground, a little over a second longer
than the one we simply dropped. How fast
will this one be traveling at the point
of impact?
Let's use this equation again. This time
we have to use 10 m/s as the initial
velocity and then we plug in our
acceleration and the time we just
calculated. This will give us negative
45.4 m/s at the time of impact, a little
more than the other rock because it fell
from a slightly higher position, the peak
of its trajectory. So we can see that
motion in the vertical direction
utilizes the same equations and concepts
as horizontal motion, except that the
acceleration will always be due to
gravity, which is held constant at this
value. Let's check comprehension.
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