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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 8

Lesson 7: Projectiles launched from/to a height# Total displacement for projectile

Reconstructing the total displacement vector for a projectile. Created by Sal Khan.

## Want to join the conversation?

- How would you know wheather or not the projectile hit the wall?(78 votes)
- Gosh! Don't over think it! You know that when it is 29.5m [30sin(80)] high, then the horizontal displacement is 5.2m [30cos(80)] if you were to divide the numbers by 3 (because I want to get 5.2m closest to 2m) I'll get vertical displacement as around 10 and horizontal displacement as 1.7-ish. So at around 2m of horizontal displacement, the vertical displacement would be a little over 10.(6 votes)

- At3:40, why do we know that displacement is 10? How can we assume that the horizontal component is large enough so that the ball will land on the ledge? It could just as easily land on the ground, making the displacement 0.(6 votes)
- You are absolutely right. He's probably worked this problem out ahead of time. But if this were on an exam and you don't know whether it reaches the cliff top or not, you could do a quick check. Check what? One approach would be to see how much time it takes to get the 2 meters to the cliff wall. Vx=30*sin(80) or 5.21m/s. So the 2 m spot is reached in 0.38seconds. OK is it high enough at that point? The y position is (at anytime it's in free fall): Y=Yo+Vyot+(1/2)at^2. Yo is 0 , Vyo is 30*sin(80)=29.5m/s and 'a' is -9.8m/s so when the ball reaches the cliff face at t=0.38sec, Y=11.2-0.7=10.5m so the projectile will be a half meter above the top of the cliff when it crosses the edge of the cliff. (If you are standing there, wear knee pads!)

It's often helpful (if time permits) to check the situation - determine a 'corner case' such as this if it's easier than the ultimate problem. What's another corner case? How about how long it takes to get to the maximum height? That happens when Vo=at. Easy to figure - but the problem is you don't know if the trajectory would run into the cliff before this happens, so this may be an unhelpful 'corner case.'(14 votes)

- does sal have video on introduction to vectors with all the laws, parrelleelgram law, product of vector,etc?(6 votes)
- Geeez, been looking at all the physics video's in search of this, didn't think of checking it at algebra. Thank you oh kind stranger(2 votes)

- So what if the base distance had been something less than 2 meters, something like 1.5 meters, in which case the projectile would crash into the side of the wall? How would you figure out vertical and horizontal displacement then? And even more challenging, what about the velocity and acceleration at the point of impact?(9 votes)
- You need to check if the tangent (relative sin and cos, or x and y direction) is greater then 10/2 (5) - which it is at 5.67(2 votes)

- realistically, wouldnt there be two times that the cannon ball passes 10m ? what if you wanted to find out the first time that the ball passes 10 m and then comes back down again to 10m?(7 votes)
- In the quadratic equation we have 2 possible values, and in this scenario, both add up to positive numbers.

t1 = 0.39s

t2 = 5.69s

This means that it takes 0.39s to get the first 10m while it's still moving upwards, then it would complete the movement and land at the platform at 5.69s.(4 votes)

- Is there ever a situation where you want a negative value on time?(3 votes)
- Yes there is. Suppose you know when the projectile hit the ground, and you want to know when it was thrown. It would not be thrown at a positive time because that would mean it will be thrown in the future. The projectile has already been thrown. The time from now when the projectile is thrown will be a negative value. If you want to know the time of something that happened in the past, you must remember that it will almost always be a negative value.(8 votes)

- In addition to understand this kind of movement, if you solve both values of the quadratic equation, you will get two "positive(+)" answers. As the teacher said, time can't and never be "negative(-)", we don't have any (-) magnitude for "time", we have 2 (+) values and the one teacher's solved is the second one which is the exact time when it reaches the landing. However, the other value of time is when it crosses 10m altitude from the ground.

Hope it was helpful, GL fellow students :)(6 votes) - How do we find out the parabolic distance covered by the object?(5 votes)
- You could use the trajectory formula. If you graph the result you'll see the entire distance traveled by the object throughout the range, or total horizontal displacement. Plugging in any values from the initial horizontal position to the final horizontal position(delta x, a.k.a. R, or range) will give you the change in height at that, and hence if you put in the total change in the horizontal position(the range) you get a graph of the vertical displacement vs the horizontal displacement. Which is basically like a map of the path the projectile takes in real life! And the path a projectile takes in space(this includes the y AND x directions, hence why you get a graph that is y vs x) is called its trajectory, and if you are wondering what its formula is, just search it up on google, it's pretty long. IF you want to know the magnitude of the total displacement vector, r, which is what it sounds like you are asking, that's even easier. For a projectile where the change in the vertical displacement is 0, thanks to no change in elevation from the starting vertical position to the final one, then the displacement vector r is equal to the magnitude of delta x, the change in horizontal displacement. Its direction is the same as that as well. For a projectile where there is a change in elevation, just apply Pythagorean theorem of the x and y displacement components and add the squares of both of them then take the square root to find the magnitude of the total displacement vector r. As for the direction, divide the y component of the displacement, delta y, by the x component of displacement, delta x, then find the inverse tangent to get the angle, which is the same as the direction, of the displacement vector r. :) so no need for calculus for that stuff.(1 vote)

- How would you find initial velocity of an object if only given 1: the height of the building it is thrown from and 2: the total displacement of the object? I'm really lost on this one....help! ex A stone thrown horizontally from the top of a 34.0-m tower hits the ground at a point 25.0 m from the base of the tower. Argg!(3 votes)
- Divide the movement into a horizontal and a vertical movement. The vertical movement is the stone dropping with gravitational acceleration for 34 m (heigth of tower). From this you can calculate the time. During the same time, the stone also moves horizontally at its inital velocity. Neglecting air friction, this velocity remains constant. So take the formula for something moving at constant velocity. You got the distance already and you get the time from solving the vertical movement equation. The result should be v0 = 13.4 m/s if I am not mistaken.(4 votes)

- Why some calculator is showing that 30 sin 80 = -29.8166?

Even Google?(2 votes)- whenever you type something like 30sin80 on google, it takes the angle in radians. you must be getting -29.81 something like that right? well, remember to specify "degrees" after you type it. then you would get the correct value, which in this case is indeed 29.54.(1 vote)

## Video transcript

- [Instructor] Let's do another example where we are projecting something and it lands at a different level and we'll also figure out
some other interesting things. We'll figure out what the
actual velocity vector is when it's landing, so both its magnitude and its direction. So let's do a situation. So let's say we are launching
something from ground level, and we're gonna launch it at a
pretty steep angle over here. So let's say we launch it with an angle, let's say we launch it with an angle of, let me just, let's say it's 80 degrees. We launch it at an 80
degree angle, at, let me. We launch it at an 80 degree angle, 80 degree angle, and
it is going to be going at 30 meters per second, 30 meters per second, so that's
the length of this vector, that's the magnitude of that vector, and let's say we want to
make it land on this landing, and this landing right over here, this landing right over here has a height, this landing over here
has a height of 10 meters. This height over here is 10 meters, so what I wanna do first of all is I wanna figure out
how far along the landing, how far along the landing
do I actually land, how far along the landing. So, and maybe I'll add some
other information right here. So say from this launching
point to the beginning of the landing, let's say that this right over here is two meters, two meters, so we just want to know how far along
the landing do we land. So like we did before, we want to break this vector into its horizontal and vertical components, its vertical and vertical components, and I'm gonna go a little
bit faster in this video, 'cause hopefully we're getting the knack for this type of thing. So our velocity, our
velocity in the vertical or our vertical component of our velocity is going to
be equal to the magnitude of our total velocity,
30 meters per second, times, and it's going to
be the sine of 80 degrees, 'cause sine is opposite over hypotenuse, times the sine of 80 degrees, and our, and we'll just get
it out of the way right now, the horizontal component of our velocity is going
to be 30 meters per second, and I'm not writing the units here, just to save space, times
the cosine of 80 degrees. Once again, cosine is adjacent, adjacent, adjacent over hypotenuse, and if you feel like I'm skipping steps, in the last few videos, I go
into this in much more detail. So, how much time do we spend in the air? How much time do we spend in the air? So, once again, in the last few videos, we saw that we can look at our, we can look at displacement, if we want to figure out time in the air, we know that displacement is
equal to the initial velocity, initial velocity times time, plus acceleration, let me
write times change in times, since that's more technically, that's technically more correct, plus acceleration times change in time squared, over, over two. Now, in our situation, we know
what our initial velocity is, we are talking about the vertical
direction right over here, so our initial velocity
is going to be this. We are trying to figure
out how time in the air, and the vertical
component determines that, because at some point, when
it hits back to the ground, it's not gonna be traveling anymore, so that's what's determining
its time in the air. So we know that, we know the acceleration. Remember, our convention, when we are dealing with
the vertical dimension is up is positive, down is negative, so this is negative 9.8
meters per second squared, and then we can, and then
what is the total displacement that we are gonna have? Well, we are starting at ground level, and we are just talking about
the vertical, remember that. So our total displacement
is going to be 10 meters, so this value right here
is going to be 10 meters, so it simplifies to 10 meters, I won't write the units here, is equals 10 is equal to,
what's 30 sine of 80 degrees? So we have 30 times the sine
of 80 degrees, gives us 29.54. 29.54. So this is 29.54 times our change in time, times our change in time, and so this is negative
9.8 divided by two, so it's negative, I'll
do that same green color, four point, that's not
the same green color, negative 4.9 meters per second squared. I'm not writing the units here, times delta t squared,
times delta t squared, and then we can subtract
10 from both sides and write this in a traditional
quadratic equation form, so we get negative, and I'm
gonna flip the sides too, negative 4.9 times delta t squared, and then we have, plus
29.54 times delta t, minus 10, minus 10 is equal to zero, and then we can use the quadratic
formula to find the roots of this, so the delta t's that satisfy this quadratic equation are going to be negative b, so negative 29.54, plus or minus the square root of 29.54 squared, b squared, minus four, minus four, times a, which is negative 4.9, for the negative times a
negative is a positive. So it's plus four times
positive 4.9, times, or I shouldn't have, I
shouldn't have jumped so fast to get rid of the negative. So it's gonna be minus four times a, which is negative 4.9, times c, which is negative 10, so just a times c, negative 4.9 times negative 10, let me just write negative 4.9 times
negative 10, these two guys, their signs are going to cancel out. All of that, all of that
over, all of that over two a. All of that over negative 4.9
times two, so negative 9.8. And like we saw in the last video, we want a positive value for this. A negative time is nonsensical, that's kind of going into the past, so we want a positive value. And since we have a
negative in the denominator, we want to have a negative value up here, and if we already have
a negative value here, if we have subtract
from that native value, we're definitely gonna have
a negative value up here, and then you divide by a negative value, we'll get a positive value, so we can really focus on
the subtracting the radical, and you could try it out. If you tried the positive version, you're gonna get a negative
value for this entire thing. Do that on, you can try
that out after this video, just to verify that you
get a nonsensical answer. So let's use a negative right over here, so we have negative 29.54
minus the square root of, the square root of 29.54 squared, and then we have minus four, minus four, times negative 4.9
times negative 10, these two, when you take the product, it's positive 49, so times, times 49, and I should add
a parentheses over there, so let me insert a parentheses. You can insert parentheses,
okay, so times 49. So this right here will
give me the numerator if I evaluate this, got me negative value, and I divide that by negative 9.8, so divided by negative 9.8
gives me 5.67, 5.67 seconds. This is equal to 5.67 seconds, and you could keep the units in there and make sure that all of the
dimensional analysis works, and I think you'll find that it does. So our total, our total vertical, the total time in the air is 5.6, the total time in
the air is 5.67 seconds. Now, what I want to do, the whole point of this is to figure out, is to figure out how far
along this landing we land. Well, our horizontal component of our velocity is right over here. We know that, we know
that our displacement in the horizontal direction, our displacement in the
horizontal direction will be our velocity in the horizontal direction, and it's a constant velocity, so it's the same thing
as our average velocity in the horizontal direction,
times the change in time. Well, the change in time is, so I'll just that times,
our change in time. So this is going to be equal
to 30 cosine of 80 degrees, 30 cosine of 80 degrees, times 5.67 seconds, times 5.67 seconds, and this, I won't write the units. This is meters per second times seconds, it'll give us an answer in meters. So once again, we have, this is our time, this is
the 5.67, times 30 cosine of 80 degrees, gives us 29.53 meters. So our total horizontal traveling, displacement I guess, we could say, is 29, I already forgot
the number, 29.53 meters. 29.53 meters, and I'll
write this as a vector, that is our horizontal displacement, it's equal to 29.53 meters. Now, we've done a lot of
deconstructing vectors, what I thought would be
interesting in this video is to construct a vector. So we know our horizontal displacement. We also know our vertical displacement, it's positive 10 meters, so
what's our total displacement? Let me write this down. So we have a horizontal displacement. We have a horizontal
displacement of 29.53 meters, and we have a vertical
displacement of plus 10 meters. We have a vertical displacement
of positive 10 meters, and so what is our total displacement? What is our total
displacement going to be? Well, we could use the
Pythagorean theorem now here. The square of the magnitude of our total displacement is
going to be equal to the sum of these two squares, or another way, this is just the Pythagorean theorem, so we could have, if we call
this length over here c, or let me just write it, this is the magnitude of our
displacement right over here, the magnitude of our total
displacement squared is going to be equal to 10
squared plus 29.53 squared. So it's going to be 10
squared plus 29.53 squared. I'm gonna do that same
color, plus 29.53 squared. If we want to solve for this, we just take the square
root of both sides, so I'll just do that in place. So if we just take the
square root of both sides, we will get the magnitude
of our total displacement, and I can get the calculator
out once again to do that. So the magnitude of our total displacement is the square root of, 10 squared is just 100, plus 29, I can even use, well, I could use all of this information, so that I don't even lose some precision. I could say second answer, that literally means the previous answer, which is that 29.53 squared, gives us a total
displacement of 31.18 meters. So this is equal to 31.18 meters, and of course, it's a vector, so we've only given you the magnitude, we also need the direction. So one way to specify direction is to give you the angle
with the horizontal, and let's call that angle theta, and once again, we can use our trig, our trig functions over here. We could use pretty much
any of the trig functions, but we know the opposite side is 10, we know the hypotenuse here is 31.18, so why not use sine, sine is opposite over hypotenuse. So we know that the sine of theta is going to be
equal to 10 over 31.18. 10 over 31.18, or if you
want to solve for theta, you take the arcsine of both sides, or the inverse sine of both sides, you get theta is equal to the inverse sine or I could write arcsine over
here, of 10, of 10 over 31.8. 10 over 31.18, I should say. So let's once again get the
calculator out to figure out that value, so I'm gonna
take the inverse sine. Once again, so this is the
same thing as the arcsine. This just says, give me the angle that when I take its sine, I get this value right over here, so the inverse sine of 10 divided by our previous answer, 31.18. I'll just say that our
previous answer is equal to, so this says, give me the angle
whose sine is 10 over 31.18. So I have 18.7 degrees, or 18.71 degrees, so this
is equal to 18.71 degrees, 18.71 degrees above the horizontal. So here we've constructed a vector, we took its vertical component and its horizontal component and we were able to figure
out the total vector, so this projectile in this situation is, its total displacement, so just to make it clear, its path will look something like this, its path will look something like this. Its path is gonna look
something like this, and we've just calculated
its total displacement, its total displacement is 31.18 meters, 18.71 degrees above the horizontal. And the other thing I've realized is when I started this problem, I asked you, I think I was asking you how far along the platform, and we've figured out its
total horizontal displacement, so if you wanna know how
far along the platform, the platform starts two
meters to the right, so it's really 27.53 meters along the platform is where it lands.