### Video Transcript

In this video, our topic is
projectile motion. Specifically, weβll be learning
some lesser known equations that describe this motion. Weβll use these to compute the
range of a given projectile, its maximum altitude, and its time-in-flight.

When we talk about projectile
motion, weβre speaking of objects that move under the influence of just one force,
gravity. So, for example, say that we have a
potato launcher that fires a potato through the air. If we neglect air resistance, then
we can say that this potato is moving only under the influence of gravity. And so, this potato is a
projectile. For one thing, that means that this
potatoβs motion is described by what are called the equations of motion. There are four such equations, and
they describe properties of our projectile, such as its final velocity, its initial
velocity, its acceleration, displacement, and time elapsed. In general, the equations of motion
can be used on any object that experiences constant acceleration. That is, as long as π is always
the same, then these equations apply. In the case of our launched potato
and any other airborne object, its acceleration is constant, and itβs the
acceleration due to gravity π.

Now, as we consider our potatoβs
overall trajectory, there are a few different quantities that we might want to
calculate. The first thing we might want to
know is how far horizontally does our potato travel. This is called the range of our
projectile. We might also want to know the
maximum altitude or elevation that our projectile achieves. We can refer to this symbolically
as β sub max. And lastly, we might want to know
the total time that our potato is in the air. Weβll call this its time-in-flight
and represent it as π‘ sub π.

It turns out that to solve for
these three quantities range, maximum height, and time-in-flight, we can use the
equations of motion. In order to do this, though, weβll
need to know something about the initial conditions of our launched projectile. Say that we take a very up-close
view of our potato launcher just as it launches this potato. Over this very short time interval,
the potatoβs trajectory would essentially be a straight line out of the
launcher. And we might be told that the
potato initially moves with a speed weβll call π£ sub π and that itβs launched at
an angle weβll call π above the horizontal. If we know these two things, the
initial speed and the initial angle of our projectile, then working with the
equations of motion, we have everything we need to solve for the range, the maximum
height, and the time-in-flight.

To see this, letβs start by solving
for π‘ sub π. Now, according to our first
equation of motion, the final velocity of our object, in this case a potato, is
equal to its initial velocity plus its acceleration times its time. If we separate out our projectileβs
motion into horizontal and vertical components, then we can apply this equation of
motion to either dimension, vertical or horizontal. Letβs say that here we focus on the
vertical component of our objectβs motion, and weβll see in a minute why we made
this choice. In that case, our equation of
motion would become the final velocity in the π¦-direction is equal to the initial
velocity of our object in the π¦-direction plus its acceleration in the π¦-dimension
times π‘ sub π.

By using this time π‘ sub π, weβre
saying that the final velocity of our projectile is its velocity just as it comes
back to Earth. As we look into its initial
velocity in the π¦-direction, its velocity just as it leaves the potato launcher, we
can indicate that using the variables in our up-close view sketch. If we let the initial trajectory of
our potato be the hypotenuse of a right triangle, then the initial vertical or
π¦-velocity of our potato is represented by this height here, and thatβs equal in
size to π£ sub π times the sin of the angle π.

So, π£ sub ππ¦ equals π£ sub π
times the sin of π, and this result can actually help us solve for π£ sub ππ¦ as
well. Hereβs how. As our potato goes through the air,
if we consider just its motion in the vertical dimension, then we can think of it as
traveling up and then coming right back down. All during its flight, itβs subject
to the acceleration due to gravity π. Since this is the only force
affecting its motion, this tells us that the speed of our projectile just as it
leaves the launcher is the same as its speed just before it lands back on Earth. On the way up, just as it goes from
its initial vertical speed and decelerates to zero, so on the way down it
accelerates and eventually reaches its initial vertical speed just before
landing. This means that when we consider
our particleβs velocity, which is a vector, we can say that the magnitude of π£ sub
ππ¦ is the same as that of π£ sub ππ¦. But the sign is opposite. In other words, π£ sub ππ¦ is
negative π£ sub π times the sin of π.

And now the only thing left to fill
in in this equation is a value for the acceleration of our projectile in the
π¦-direction. But we already know that thatβs π,
the acceleration due to gravity. Hereβs something important to keep
in mind, though. Because we defined vertically up as
the positive π¦-direction, this means our acceleration in the π¦-direction will have
to be negative since it acts downward. So, if π is 9.8 meters per second
squared, then π sub π¦ must be negative π.

Note that now weβre at a point
where if we know π£ sub π and π, we can rearrange this equation to solve for π‘
sub π. If we first subtract π£ sub π
times the sin of π from both sides, then we get negative two π£ sub π sin π
equals negative π times π‘ sub π. Dividing both sides by negative π,
that factor cancels on the right and the negative signs cancel on the left. And so, we find that two times π£
sub π times the sin of π over π equals π‘ sub π, the time-in-flight of our
projectile.

Letβs record this result off to the
right. And now, letβs work on solving for
the maximum height that our projectile achieves. Once again, this has to do with the
vertical rather than the horizontal component of our objectβs motion. The equation of motion weβll choose
to work with is this second one here. We choose this because it has the
variable we want to solve for β remember that π represents displacement β while we
know the values for the other variables involved, in this case the vertical
acceleration, the initial vertical velocity, and final vertical velocity.

Regarding that final vertical
velocity of our object, weβve called it π£ sub ππ¦, we know that when our potato is
at its maximum height, its position will be right here and that at that moment in
time, its vertical velocity will be zero. When the potato is at its high
point then, its velocity in the π¦-direction is zero. So, the left-hand side of this
equation becomes zero. Just like before, the potatoβs
initial velocity in the π¦-direction is π£ sub π times the sin of π and its
acceleration in the π¦-direction is negative π. This gives us this equation. And recall itβs β max we want to
solve for.

What weβll do then is subtract π£
sub π sin π quantity squared from both sides. And from there, we divide both
sides of our equation by negative two times π. This cancels that factor on the
right-hand side and eliminates the negative signs on the left. So, in terms of the initial speed
and direction of our projectile, the maximum height it will reach is equal to the
quantity π£ sub π times the sin of π squared all over two times π. And note that another way to write
this is π£ sub π squared times the sin squared of π over two π.

Okay, so weβve covered
time-in-flight and maximum height achieved. Now, letβs look at how to calculate
the range of a projectile. Weβll do this a bit differently
than before, because now the horizontal velocity of our projectile is important. Looking over at our initial
zoomed-in view, we know that the initial horizontal velocity of our projectile is π£
sub π times the cos of π. And even though weβve called this
the initial horizontal velocity, this actually doesnβt change over time. Because thereβs no acceleration on
our projectile in the horizontal direction, its horizontal velocity is constant. This is why earlier on, we chose to
use the vertical dimension in looking at this equation of motion rather than the
horizontal.

In a horizontal plane, the
acceleration is zero. So, we just have an equation that
says π£ sub π times the cos of π equals π£ sub π times the cos of π. Thatβs not so helpful. And so, thatβs why we didnβt choose
that dimension. But now, we do need to work with
velocity in this direction because we can see it will have an impact on the range
that our projectile travels. To start figuring this out, letβs
choose an equation of motion that will suit our needs.

First, we want to use an equation
of motion that involves the variable π ; that includes three of these four. Second, it will be nice to take
advantage of the fact that in the horizontal direction π is zero; in other words,
any term with π in it will go to zero. That leaves us with this equation
of motion and this one. Looking more closely, though, we
can see that actually, this equation of motion isnβt an option for us. Because if π is zero, then that
cancels out the term that involves the factor we want to solve for, π . That settles it then; weβll work
with this equation of motion to solve for the range.

The equation we can write is that
the range of our projectile is equal to its initial velocity in the π₯-direction
times the total time-in-flight π‘ sub π plus one-half its acceleration in the
π₯-direction times the time-in-flight squared. As we mentioned, π sub π₯ is zero
because thereβs no acceleration in the horizontal direction. Weβve also seen that π£ sub π in
the π₯-direction is π£ sub π times the cos of π. And then for π‘ sub π, we can
refer to the equation that we derived earlier. Itβs two times π£ sub π times the
sin of π over π. So, the range our projectile
travels is equal to this product, which is equal to two times π£ sub π squared
times the cos of π times the sin of π all over π.

So then, we now have equations for
the range, maximum height, and time-in-flight for a projectile given its initial
velocity, that is, its initial speed and the direction of travel. Knowing all this, letβs get a bit
of practice now through an example.

A projectile has an initial speed
of 25 meters per second and is fired at an angle of 48 degrees above the
horizontal. What is the time between the
projectile leaving the ground and returning to the ground at the same height that it
was launched from?

Okay, letβs say that this is the
path that our projectile follows. Weβre told that when itβs initially
launched, it has a speed of 25 meters per second and that itβs fired off at an angle
weβll call π of 48 degrees. Knowing this, we want to solve for
the time it takes for this projectile to return to the surface of the Earth. In other words, how long does its
full trajectory take? Weβll call this time π‘ sub π. And since weβre working with a
projectile, that means we can recall the equation for this time.

Given a moving object affected only
by the force of gravity, if that object is launched from the ground with an initial
speed π£ sub π at an angle π, then its total time-in-flight is given by this
expression. Since we know our objectβs initial
speed and its angle of launch, and we also know that π is equal to 9.8 meters per
second squared, we can plug these values into our equation for π‘ sub π. When we enter this expression on
our calculator, to two significant figures, we find the result of 3.8 seconds. Thatβs the total time that this
projectile is in flight.

Letβs look now at another example
exercise.

A projectile is fired at an angle
of 32 degrees above the horizontal with an initial speed of 44 meters per
second. What is the maximum upward vertical
displacement of the projectile from its launch position?

Okay, so, letβs say that this is
ground level and this orange line shows us the trajectory of this projectile. The projectile is fired at an
initial speed weβll call π£ sub π and at an angle weβll call π. We want to solve for the maximum
upward vertical displacement the projectile reaches, that is, this displacement
here. Weβll call it β sub max.

Since weβre working with a
projectile and we know its initial speed and direction of travel, we can recall an
equation for the maximum height reached by a projectile in terms of its initial
speed and its direction. Since weβre given π£ sub π and π
and we also know that the acceleration due to gravity is 9.8 meters per second
squared, we can plug these values into our equation for β sub max. When we enter this expression on
our calculator, to two significant figures, we find a result of 28 meters. This is the maximum vertically
upward displacement of our projectile.

Letβs look now at one last example
exercise.

A projectile is fired at an angle
of 66 degrees above the horizontal. The time between the projectile
leaving the ground and returning to the ground at the same height that it was
launched from is 2.9 seconds. What was the projectileβs Initial
speed?

Okay, letβs say that this is the
path that our projectile follows. The initial launch angle of this
projectile, weβll call it π, is 66 degrees. And the time it takes for our
projectile to travel this full path, weβll call this π‘ sub π, is 2.9 seconds. Knowing this, we want to solve for
the initial speed of the projectile. Weβll call it π£ sub π. Since we are working with a
projectile, a body moving under the influence only of the force of gravity, we can
recall that the total time it takes for a trajectory that starts and ends at the
same height is equal to two times the projectileβs initial speed multiplied by the
sin of its launch angle all divided by π.

In our situation, itβs not π‘ sub
π we want to solve for but π£ sub π. To begin doing that, we can
multiply both sides of this equation by π over two times the sin of π. Over on the right-hand side, the
factors of two, π, and sin π all cancel out, leaving us with π£ sub π. And we see then that π£ sub π is
equal to π times π‘ sub π over two times the sin of π.

Looking at whatβs given to us in
our problem statement, we know π as well as π‘ sub π, and we can recall further
that π equals 9.8 meters per second squared. If we then plug in these values to
our equation for π£ sub π, to two significant figures, we get a result of 16 meters
per second. This is the initial speed our
projectile would need to have at this given launch angle so that itβs in the air for
2.9 seconds.

Letβs now summarize what weβve
learned in this lesson through a few key points. We saw here that for a projectile
that lands at the same height from which it was launched, if we know its initial
launch speed π£ sub π and its initial launch angle π, then its total
time-in-flight π‘ sub π equals two times its initial speed times the sin of π
divided by π. Similarly, the maximum height above
ground level that it achieves, β sub max, equals its initial speed squared times the
sin of this angle π squared divided by two times π. And lastly, the projectileβs range,
its total horizontal distance traveled, is equal to two times π£ sub π squared
times sin π times cos π all over π.