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## Physics library

### Course: Physics library > Unit 2

Lesson 1: Two-dimensional projectile motion- Horizontally launched projectile
- What is 2D projectile motion?
- Visualizing vectors in 2 dimensions
- Projectile at an angle
- Launching and landing on different elevations
- Total displacement for projectile
- Total final velocity for projectile
- Correction to total final velocity for projectile
- Projectile on an incline
- 2D projectile motion: Identifying graphs for projectiles
- 2D projectile motion: Vectors and comparing multiple trajectories
- What are velocity components?
- Unit vectors and engineering notation
- Unit vector notation
- Unit vector notation (part 2)
- Projectile motion with ordered set notation

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# Launching and landing on different elevations

More complicated example involving launching and landing at different elevations. Created by Sal Khan.

## Want to join the conversation?

- I understand that time can't be negative for the purposes of this video, but can it be negative in any other situation? Please specify an example.

thanks(54 votes)- The "time" in these videos is always the "change in time". So yes, it is quite logical to have a negative change in time. It would be before some event, such as before the projectile was launched.(92 votes)

- My only issue with this video is that you didn't specify the length of the 'well' in the middle. How do I know that the projectile won't land there instead of the second plateau?(42 votes)
- In other words, we are told that the projectile lands on the second plateau. That is a fact. What we could ultimately answer is the magnitude of the horizontal displacement (i.e. the distance from the cannon to the place where the projectile lands). The length of the 'well' will remain forever a mystery!(27 votes)

- @15:12, why didn't Sal use the formula: s = Vi(x) + a(deltaT)

is it because the acceleration -9.8m/s^2 only applies in the vertical direction?

I'm just confused because I thought V(x) of 90cos53 was only for Vi(x) and not Vavg(x)

Thanks for answering!(8 votes)- Sal, worked this problem in such a way to avoid using several other algebraic equations. He knows that Juniors and Seniors in high school and college know the quadratic equation. So he solved for the time by this means. There is another way (using basic algebra....takes forever)...For Y(direction)....You will need to use....Vf=Vi+at (Vf=0 Vi=90sin53 a=-9.8) t=7.33(this is just the time up) but total air time would be (7.3344*2= 14.67s)..to land on its original horizontal axis...and a=-9.8 because the object is going up and gravity is slowing its upward velocity...to where Vf=0 at 7.3344s)....Now as the objects falls, since the y-direction changes...its as if Vi=0 and Vf=90sin53...as it returns to its horizontal reference...BUT once it falls below it....we have a new calculation....to consider....Our Vi=90sin53 because it has an initial velocity but we now do not know its new final velocity...We have to find it.... Vf^2=Vi^2+2ad (Vi=90sin53 and a=9.8 and d=16) so Vf=74.03m/s..Now its is time to find how long it took to fall this 16m.......then Vf=Vi+at (Vf=74.03 Vi=71.88 a=9.8)and a=+9.8 because gravity is helping its Velocity increase....thus....t=0.22s (time it takes to fall the 16m)....final time in the air will be 14.67s+0.22= 14.89s....then to get the X-distance just use D=V*t...(D=90cos53*14.89= 806.5m....Sorry for the long winded explanation.(20 votes)

- Very stupid question:

Is the horizontal component of a velocity vector, how far the object was displaced or the total time it was in the air?(5 votes)- It is the velocity in the horizontal direction.(15 votes)

- how come all the equations used are only applied to the vertical velocity of the cannonball?(5 votes)
- hi, so does this mean that time up = time down also applies here when the elevation is different? sorry and thanks!(2 votes)
- No. All Sal did was find the total time that the object was in the air. Time up=Time down does not apply.

Also watch the next video.

for further clarification - informjaka@gmail.com

hope it helped(7 votes)

- Sal has solved this using Quadratic formula, but I don't understand it is there any other way solve it?

Like by factoring or something?(4 votes)- u can use it,

but takes time.

the formula is faster and easier(2 votes)

- hello . i want to ask why did sal put 90sin53 as a initial velocity instead of 90m/s , can someone explain it more?(4 votes)
- how come 2*negativeA gives another negative value11:30(3 votes)
- Any positive times a negative gives a negative.(5 votes)

- Hi, I don't understand why the horizontal velocity is constant. Is it always this case? Thank you.(3 votes)
- Horizontal velocity is always constant, when the air resistance is negligible. With an acceleration, the component of velocity cannot change. Vertical velocity is affected by gravitational pull (9.8m/s). But the horizontal velocity, faces zero negative or positive acceleration in the mid air, until you are talking of a self-accelerating body like flight. So generally, the horizontal velocity is taken constant.(3 votes)

## Video transcript

Let's do a slightly
more complicated two-dimensional projectile
motion problem now. So in this situation, I am going
to launch the projectile off of a platform. And then it is going to
land on another platform. And I'm going to fire the
projectile at an angle. Let me draw this a
little bit better. So I'm going to
fire the projectile at an angle of-- let me
use a not so clean number. Let's say it is a
53-degree angle. And it's coming
out of the cannon. And let me make it clear. So it's coming out--
let me do it this way, just to make it 100% clear. So this angle right
over here is 53 degrees. And we are going
to have it come out of the muzzle of the cannon
with a velocity of 90 meters per second. And just to give ourselves
a sense of the heights, or how high it's being
launched from-- so from the muzzle of the
cannon down to here. So this height right
over here, let's say that that is 25 meters. And let's say that this height
right over here is 9 meters. And so we're essentially
launching this from a height of 25 meters. I know in the last video,
even though I drew the cannon like this, we
assumed that it was being launched from
an altitude of 0 and then landing back
at an altitude of 0. Here we're assuming
we're launching it from an altitude of 25
meters, because that's when it's leaving the muzzle. And it's going to
start decelerating at least in the
vertical direction as soon as it leaves the muzzle. And then we're
assuming it's not going to land back at
the same altitude. It's going to land at
a different altitude. So how do we think
about this problem? So the first thing
you'll always want to do is divide your velocity
vector into its horizontal and vertical components. You use the vertical
component to figure out how long it's going
to stay in the air. And then you use the horizontal
component to figure out, given how long it's in the
air, how far did it travel. And once again,
we're going to assume that air resistance
is negligible. So just based on what we
did in the last video-- and I'll go through all of the
steps in this one, as well. So if we draw our vector, the
length here is going to be 90. This is our velocity vector. The angle over here between
the x-axis and our vector is 53 degrees. And let me draw the
horizontal component. The horizontal component
would look like this. And the vertical component--
I'll do it in orange. The vertical component would
look like-- that's not orange. The vertical component
will look like this. And so the vertical
component of the vector-- what would be the length of
this side right over here? Well, this is the opposite side. We know from basic
trigonometry sine of an angle is opposite
over the hypotenuse. So we know that the
sine of 53 degrees is equal to this
opposite side, is equal to the vertical velocity. And I could write
it's the magnitude of the vertical velocity. I write that subscript
y, because we're in the y direction. That's the vertical
direction, over the length of the hypotenuse,
over the magnitude of our original vector. Or we can get that this
side right over here. If we multiply both
sides by 90, we get that the
magnitude of that side is going to be equal to 90
times the sine of 53 degrees. Now, if we want to do
the horizontal component, the horizontal side
is adjacent to this. Cosine, soh cah toa. Cosine is adjacent
over hypotenuse. So the horizontal
component of our velocity, I'll say in the x direction,
over the hypotenuse, over 90, is equal to the
cosine of 53 degrees. Cosine is adjacent
over hypotenuse. Adjacent, that's
this length, over 90. Multiply both sides by 90. You get that the
horizontal component is equal to 90 times
cosine of 53 degrees. Now, how do we
figure out how long this thing stays in the air? Well, we'll use the
vertical component for that. And especially since we're
dealing with different levels, we can't use that more basic
reasoning, that hey, whatever velocity we start
off at, it's going to be the same velocity but
in the opposite direction. Or the same magnitude
and velocity but the opposite direction. Because we're not going
to the same elevation. But what we could do is
we can use the formula that we derived in
the previous video. That the displacement--
let me just copy and paste this a little bit
lower so we can use it. So I'll stick it
right over here. So we could use this. We know that the displacement is
equal to the initial velocity-- and we're dealing with
the vertical direction right here-- times the change
in time plus the acceleration, times the change in time
squared, divided by 2. So how do we use this to figure
out how long were in the air? So what is the displacement? If we're starting
at 25 meters high, and we're going
to 9 meters high. So over the course, while
this thing is traveling, it will be displaced
downwards 16 meters. Or another way to
think about it is our displacement in
the vertical direction is going to be equal
to negative 16 meters. Let me write that a
little bit bigger. Negative 16 meters, right? Because 25 minus 9 is 16. And so we can put
that into the formula that we derived in
the previous video. We get negative 16-- I won't
write the units here just so that we don't take up
too much real estate, so that at least
it looks simple-- is equal to the
initial velocity. We're dealing with just the
vertical dimension here. So we're just dealing
with the vertical. And remember, it's negative
because our displacement is going to be downwards. We're losing altitude. So our vertical velocity--
we already figured that out. It is 90 times the
sine of 53 degrees. Actually, let me do
it in that same color. The first time we do
a new type of problem, it's good to know-- so 90
times the sine of 53 degrees, times our change in time,
is equal to the acceleration of-- due to the force of gravity
for an object in free fall is going to be negative 9.8
meters per second squared. But we're dividing that by 2. So we're going to have minus 4.9
meters per second squared times delta t squared. Times our change
in time squared. So how do we solve
something like this? You can't just factor
out a t and solve it. But you might recognize
this is a quadratic equation right over here. And the way you solve
quadratic equations is you get everything onto
one side of this equation. And then you either
factor it out. But more likely
in this situation, we will use the
quadratic formula, which we've proved
in other videos and hopefully given you the
intuition for it-- to actually solve for the times where
your elevation, where your displacement in
the vertical direction is negative 16 meters. And you'll get two
solutions here. And one of the solutions will
be a negative change in time. So it'll be like at
sometime in the past, you were also at
negative 16 meters. That's nonsensical
for this problem. So we'll want to take
the positive value here. So let's put all of this on
one side of the equation. So let's add 16 to both sides. On the left-hand side,
you just get a 0. 0 is equal to--
and I'll write it in kind of the traditional way
that we're used to seeing it. I'll write the highest
degree term first. So negative 4.9 times
delta t squared. And then we have plus 90 sine
of 53 degrees times delta t, and then plus 16. I'm going to do that in yellow. All of this is equal to 0. And this, once again, is
just a quadratic equation. We can find its roots. And the roots will be
in terms of delta t. We can solve for delta t
using the quadratic formula. So we get delta t-- and if
this is very unfamiliar to you, review the videos on Khan
Academy in the algebra playlist on the quadratic formula. And if you don't know
where it came from, we also prove it for you. So it's equal to negative B.
B is this term right here, the coefficient on the delta t. So it's going to be negative
90 sine of 53 degrees. I'll write the quadratic
formula up here for those of you who
don't fully remember it. So if I'm trying to
solve Ax squared plus Bx plus C is equal to 0,
the roots over here are going to be negative B plus
or minus the square root of B squared minus 4AC, all
of that over 2 times A. These are going to
be the x values that satisfy this equation up here. So that's all I'm
doing over here. This is the B value. Negative B plus or
minus-- and it's going to turn out that we
only care about the plus one, because that's going to
give us the positive value. But I'll just write it out here. Plus or minus the square
root of B squared. So it's this quantity squared. So it is 90 sine of 53
degrees squared, minus 4-- we're going to run
[? out that ?] little space. So minus 4 times A,
which is negative 4.9. Well, let me just write negative
4.9 times C. C over here is 16, times 16. Let me put this radical
all the way over here. All of that over 2A. So A is negative 4.9. 2 times A is negative 9.8. So now we can get
the calculator out to figure out our
change in time. And I'm just going to focus
on the positive version of it. I'll leave it up to you to
find the negative version and see that it'll give you
a negative value for change in time. And that's
nonsensical, so we only care about the
positive change in time where we get to a displacement
of negative 16 meters. Let's get the calculator out. So we get-- want to
do this carefully. We have negative 90
sine of 53 degrees plus. I'm doing the plus version
here because that'll give us the positive value--
plus the square root of. And I'll do this in parentheses. 90 sine of 53 degrees squared. That's that part right there. These two negatives cancel out. So I could say this is
plus 4 times positive 4.9. So plus 4 times 4.9 times 16. And then that closes
off our entire radical. And so this will give me
the numerator up here. That gives me the numerator. And then I want to divide that
by-- did I do the negative 90? Oh, and I just realized
that I made a mistake. I said that the positive
version would give you the positive time. But now we realize that's wrong. Because when I took the positive
version, when I put a plus up here, I get a positive
2.14 for the numerator. But then we divide
it by negative 9.8. We're going to get
a negative value. So that's not going to be
the time that we care about. We care about the time where
this is a negative value. So let me re-enter that. So let me do the negative value. So let me move
back a little bit. And then let me replace
this with a minus sign. So I'm going to look
at the negative value, because I want
the positive time. And so now my numerator
here is a negative value. And so this is actually
what we care about. We care about the
numerator's a negative value. You divide by
negative 9.8, and you get 14.8-- I'll just
round-- 14.89 seconds. So delta t, the
positive version, is equal to 14.89 seconds. And so my initial comment about
using the positive version was wrong because we have this
denominator that's negative. So you want the
numerator to be negative. And only when the
numerator is negative will the whole
expression be positive. And so we got this positive
time of 14.89 seconds. I'm going to leave you there. And the next part of
the video-- actually, I might as well just solve it
instead of making a new video. Although this is running long. So the amount of time that we're
in the air is 14.89 seconds. So if I were to ask you the
horizontal displacement, it's going to be
the amount of time we're in the air times your
constant horizontal velocity. And we already figured out our
constant horizontal velocity. So if you want to figure
out how far along the x-axis we get displaced, we
just take this time times-- that just means
our previous answer-- times this value right here,
times 90 cosine of 53 degrees. And that gives 806 meters. So this displacement right
over here is 806 meters.