It's professor Dave, let's learn about
projectile motion.
We've learned how to use kinematic equations
to describe the motion of an object
moving horizontally, like a car, as well
as objects moving vertically, like
objects falling straight down to the
ground, but what about objects that move
in both of these directions? This type of
motion, if it involves an object that is
thrown or launched into the air, can be
referred to as a projectile motion.
Imagine a cannonball being fired at some
angle from the horizontal. It will travel
some distance up into the air before
eventually falling back down and hitting
the ground, some distance away from the
cannon, and we can use a parabola to
represent the path of this object.
The important thing to understand about
these kinds of examples is that the
horizontal motion and vertical motion of
the cannonball are completely
independent of one another.
This means we can use separate equations
to discuss the motion in each direction,
one equation that exclusively
corresponds to the x-coordinates of the
object and another that exclusively
corresponds to the y-coordinates of the
object. To drive this idea home consider
two marbles, one dropped from a
particular height and another that rolls
off of a surface at that same height
with some horizontal velocity. If these
begin falling at the same time they will
strike the ground at the same instant
because their vertical motion is
independent of any horizontal motion. The
one with horizontal velocity will cover
some distance in the X direction but it
will fall downward at the same rate as
the one that falls straight down and so
they will have identical airtimes. How
can we apply this to other real-world
examples?
well we have been throwing lots of rocks
off of cliffs lately so let's throw one
more. This time let's throw a rock at an
upward angle of 30 degrees off the
horizontal from the very edge of our
favorite 100-meter cliff and with an
initial velocity of
8.5 meters per second. Again we will ask
how long before the rock hits the ground?
and now additionally we want to know how
far away from the edge of the cliff it
will land. First things first let's make
sure we understand how these questions
relate to motion in both the x and y
direction. The time it spends in the air
only relates to Y direction behavior
because it will stop being in the air
when it hits the ground, no matter what
the horizontal velocity is, from 0 to
some huge number.
The vertical motion will be independent
of that. The distance it travels from the
edge of the cliff depends on the
horizontal velocity but also the time it
spends in the air because once it hits
the ground it can't travel any further.
And the velocity at any moment can also
be split into components. The horizontal
velocity will be the same at every
moment in this trajectory as long as we
disregard wind resistance, but the
vertical velocity will be the greatest
at the moment the rock is thrown and
then decrease until it reaches zero at
the zenith and then become increasingly
negative until it hits the ground.
This is because there is a constant
acceleration in the negative direction
due to gravity. We know we can use these
equations to answer these questions but
in order to discuss the x and y
directions
separately we have to split up this
velocity vector into x and y components.
This is pretty simple to do. We can
simply draw horizontal and vertical
vectors starting from the same point as
this one and extending precisely as far
as this one does in the X or Y direction.
These can be labeled V sub X and V sub Y
and we can think of these as being the
legs of the right triangle with the
existing vector as the hypotenuse. To
find out the magnitude of these vectors
we just do a little trig. We know that
the cosine of 30 will be equal to the
adjacent leg over the hypotenuse which
means 8.5 cosine 30 or 7.36 will be the
initial
velocity in the X direction.
We also know that the sine of 30 will be
equal to the opposite leg over the
hypotenuse so 8.5 sign 30 or 4.25
will be the initial velocity in
the y-direction. These velocities are
independent of one another, meaning that
a rock that was thrown straight up with
an initial velocity 4.25 m/s will land
on the ground at precisely the same time
as the one thrown at 8.5 meters per
second but at this 30 degree angle. That
means that if we want to figure out how
long it is in the air
this is no different than a problem with
one-dimensional vertical motion provided
that we understand that we are only
looking at the y component of this
velocity vector. So we can plug the
numbers in, -100 for the
displacement, 4.25 for initial velocity
and our usual acceleration. Putting it
into standard form and plugging into the
quadratic equation we will take the
positive result for t and say that the
rock will be in the air for 4.97 seconds.
This also makes it easy to calculate how
far the rock will travel, because we know
the horizontal velocity will be 7.36 m/s
times the 4.97 seconds that it is in the
air which leaves us about 36.6 meters
away from the edge of the cliff
So projectile motion is a little more
complicated than one-dimensional motion
but if we know how to divide any vector
into x and y components we can easily
analyze projectile motion as the
combination of two types of
one-dimensional motion. Let's check comprehension.
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