Hello everyone. I'm Jesse Mason and in this installment of Teach Me, we're going to take
a look at a classic projectile motion problem involving an angled launch from uneven ground.
For this problem we're given the initial position and velocity of the projectile and our task
is to determine the projectile's range, peak height and velocity at some point in time
along its trajectory. The first step, as always, is to draw a picture.
So here's the cliff from which our projectile is launched; and here's the level ground below.
We'll draw a cannon to fire the projectile and also to mark's its initial position. (Loud
boom!) And there's our trajectory. "Take that, Pachelbel!"
And we're done with the picture. Not much to this one... yet.
So what are our knowns-n-unknowns? To start, we're given the height of the cliff, which
is to say the initial vertical displacement. That's one hundred meters. Of course, we've
already implied our coordinate system but we need to be explicit and include it on our
drawing. Next, we're given the magnitude of the launch velocity -- the initial speed -- of
our projectile: thirty meters per second. We're also given the direction of the launch
velocity, so we'll label the angle between the velocity vector and the horizontal as
theta. Theta is equal to twenty degrees.
So there are some of our knowns. There are a few more that are implied and we'll address
them shortly.
The first unknown we're tasked with determining is the projectile's horizontal displacement
upon impact: its range. We'll label this x. And by doing so, we've implied the location
of the origin so let's indicate that on our picture. NOW it should be clear to see that
our projectile's initial position along the x-axis is zero meters. We'll label the maximum
vertical displacement -- that is to say, our peak height - as y-max.
Lastly, we're looking for the projectile's velocity two seconds after launch. We're gonna
guess that the projectile -- a cannonball in this case - would be somewhere around...
here... at that time. Accuracy with our guess is unimportant because the equations will
tell us... the rest of the story. So at this point in the trajectory the instantaneous
velocity looks something like this. We're after the magnitude and the direction so we'll
label the angle between the velocity and the horizontal as phi. (Fee? Fi? Fee fi fo fum,
whatever. It's all Greek to me.) Annnnd let's not forget to jot down the given time for
this part. Since we'll be looking for each component of the projectile's velocity separately,
we'll break v into v-sub-x and v-sub-y. And while we're at it, let's resolve our initial
velocity into its constituent x- and y-components as well. So here's our given launch velocity,
v-sub-oh. In the vertical direction we have v-sub-oh-sub-y. And in the horizontal we have
v-sub-oh-sub-x and here's the given launch angle theta. Since we'll eventually be writing
these components in terms of v-sub-oh and theta, we'll write their trigonometric relationship
right here. Let's see... so the y-component of v-sub-oh is opposite theta, so it's equal
to v-sub-oh sine theta. And the x-component of v-sub-oh is adjacent to theta so it's equal
to v-sub-oh cosine theta.
By the way, we've got an implied known that should be listed: the acceleration due to
gravity. What value should we use for g? Or more to the point, is the magnitude of g positive
or negative? Well, it all depends on our coordinate system so we need to compare the acceleration
vector to our chosen coordinate system. What's the direction of the acceleration vector?
Downward, of course. Since we aligned the positive y-axis with the upward direction,
and the acceleration vector is downward, that means that the magnitude of g for this problem
will be negative 9.8 meters per second squared.
Okay! We're finished with identifying the knowns and unknowns. Now we're ready to move
on to the next step: selecting the appropriate equations for our problem.
Since we're dealing with projectiles and ignoring the effects of air resistance, the kinematic
equations -- namely, the displacement equation and the velocity equation -- are entirely
sufficient to analyze the motion of our projectile. Of course, our projectile is moving through
two dimensions so -- with a nod of appreciation to Galileo -- we'll apply these equations
to each direction separately.
So in the x-direction, starting with the displacement equation: Instead of s, we have x equals one
half a-sub-x (that's the acceleration in the x-direction) times t squared, plus v-sub-oh-sub-x
(the x-component of the initial velocity) times t, plus x-sub-oh (the initial horizontal
position). Before moving on, we're going to simplify this equation. A projectile's horizontal
motion is unaccelerated so we can set a-sub-x to zero. We've defined v-sub-oh-sub-x as v-sub-oh
cosine theta. And x-sub-oh is zero. So the displacement equation in the x-direction simplifies
to: x equals v-sub-oh cosine theta t. We'll label this equation (1) and set it aside for
now.
Now we apply the velocity equation to the x-direction and we get: v-sub-x equals v-sub-oh-sub-x,
plus a-sub-x times t. We'll set v-sub-oh-sub-x to v-sub-oh cosine theta. And again, the horizontal
acceleration of a projectile is zero. So the velocity equation in the x-direction simplifies
to v-sub-x equals v-sub-oh cosine theta. We'll label this equation (2) and set it aside.
Okay, so now we'll apply the kinematic equations to the y-direction. The displacement equation
becomes: y equals one half a-sub-y t squared, plus v-sub-oh-sub-y t, plus y-sub-oh. First,
a-sub-y, the acceleration in the y-direction, is the gravitational acceleration of our projectile
so we can set that to g. And we've defined v-sub-oh-sub-y (the y-component of our projectile's
initial velocity) as v-sub-oh sine theta. So the displacement equation in the y-direction
simplifies to become: y equals one half g t squared, plus v-sub-oh sine theta t, plus
y-sub-oh. Label that equation (3) and we're almost done here.
The velocity equation in the y-direction is v-sub-y equals v-sub-oh-sub-y, plus a-sub-y
times t. Again, v-sub-oh-sub-y was defined as v-sub-oh sine theta and a-sub-y is just
g. So our fourth and final equation is v-sub-y equals v-sub-oh sine theta plus g t.
These four equations now constitute what we call our toolbox. Every handyman knows you've
gotta have the right tools for the right job. The same goes for projectile motion: simplifying
the displacement and velocity equations for each direction is selecting the right tools.
And with these tools we can determine everything we want to know about our projectile's motion
and trajectory.
First up: determining the projectile's horizontal displacement upon impact, i.e., its range.
Where do we start? Well, equation (1) would give us horizontal displacement but we don't
have t, the time of impact. To get t, we're going to need to use equation (3) along with
a key implication: at the moment of impact, the projectile's vertical displacement is...
zero! Now we can solve equation (3) for time.
So we'll start by writing equation (3). We'll set y to zero and rewrite for clarity's sake.
Does this equation ring any algebraic bells? Perhaps if we recall it's more general form
and solution: If zero equals A t squared plus B t plus C (where A, B, and C are all constants),
then t equals negative B plus-or-minus the square root of B squared minus four A C all
over two A. Yep, that's how I remember the quadratic formula. So our one half g here
will play the role of A, our v-sub-oh sine theta will be B, and our y-sub-oh is C. So...
t equals (long inhale) negative v-sub-oh sine theta plus-or-minus the square root of quantity
v-sub-oh sine theta end quantity squared minus four times one half g y-sub-oh all divided
by 2 times one half g. (Applause) Thank ya, thank ya very much. Simplifying just a few
terms and then inserting our values we find that t equals negative 3.59 seconds or positive
5.68 seconds. The negative value is not useful for this problem -- negative time? what does
it mean? -- so we'll lovingly discard it.
Now we proceed with equation (1) -- write it out here -- and insert the appropriate
values: 30 meters per second, cosine twenty degrees and for time we use our 5.68 seconds
from equation (3). And we get a horizontal displacement of 160.1 meters. All things considered,
not an unreasonable range.
Okay, now let's determine the maximum vertical displacement of our projectile, that is, the
peak height of its trajectory. We'll use equation (3) to determine this value but we first need
to obtain the time which corresponds to this moment. To obtain the time we need to recognize
that the moment our projectile crests its trajectory, the vertical component of its
velocity is zero. Using this value, along with equation (4) will give us the time at
which our projectile reaches its peak height.
Okay, solving equation (4) for time, we get t equals negative v-sub-oh sine theta divided
by g. Inserting our values, we find that at 1.05 seconds our projectile levels out and
begins its descent, so to speak. Now we'll insert this time into equation (3) to determine
the maximum vertical displacement of our projectile. And when we assign our values to the variables
-- using 1.05 seconds, the time coincident with y-max, for t - we find that our projectile
reaches a peak height of 105.4 meters. Not very impressive, but very reasonable given
our shallow launch angle.
Lastly, we'll determine the projectile's velocity 2 seconds after launch. We'll need to determine
the x- and y-components separately so starting with v-sub-x, we'll use equation (2). V-sub-x
equals v-sub-oh cosine theta. Using the aforementioned known values we get 28.19 meters per second
for the horizontal component. We'll use equation (4) to determine the vertical component. Inserting
our known values, including the time in question -- two seconds - we get a value of negative
9.34 meters per second for v-sub-y. What's up with the negative sign? The negative sign
implies motion in the negative y-direction, that is, downward.
Now at this point we could write our solution in vector notation, where the velocity vector
is equal to the magnitude of the velocity's x-component times the unit vector i-hat, plus
the magnitude of the velocity's y-component times the unit vector j-hat. That would give
us the solution v equals 28.19 meters per second i-hat, plus negative 9.34 meters per
second j-hat... which is fine and dandy, but let's go ahead and determine the magnitude
and direction of this velocity vector.
So we've got 28.19 meters per second in the x-direction and 9.34 meters per second in
the negative y-direction. We'll recombine these components to determine the magnitude
of v using Euclid's 47th Proposition, better known as Pythagoras' Theorem: the square of
the hypotenuse is equal to the sum of the squares of the legs. Taking the square root
of both sides and inserting our values (and dropping the plus-or-minus sign) we get a
magnitude of 29.70 meters per second. Physically reasonable? Check. Are we done? Not yet, we
still need the direction of its motion, that is to say the angle between the velocity vector
and the horizontal. To get phi, we use trigonometry. So the tangent of phi equals opposite (9.34
meters per second) over adjacent (28.19 meters per second). Take an inverse tangent and we
get a value of 18.3 degrees for phi.
So two seconds after launch our projectile is moving at 29.70 meters per second at 18.3
degrees below the horizontal.
I'm Jesse Mason. I hope you found this video helpful and that I've convinced of the utility
-- nay, the necessity! -- of creating a toolbox to solve projectile motion problems. If you
have any suggestions for future Teach Me videos or just want to say hello from your part of
the world, please do so in the comments section below. And as always, happy learning!