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## Class 11 Physics (India)

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

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

Calculating the total final velocity for a projectile landing at a different altitude . Created by Sal Khan.

## Want to join the conversation?

- isn't the final vertical velocity is -26.03 instead of -29.03??(16 votes)
- Yes, Sal mixed them up. It should be 26.03, and the final degree measure is 78.7 degrees. There will be a correction in the next video.(7 votes)

- How would you do this using calculus?(12 votes)
- Can someone explain why the horizontal velocity haven't changed due to acceleration ? Why is acceleration 0 on the horizontal line ?(4 votes)
- Acceleration is zero on the horizontal because a few videos ago,we actually assume that drag is negligible. So, if drag is neglible and we ignore friction as well, there are no forces acting upon it. If there are no forces acting upon something then, the acceleration is zero.

This can be seen by f=ma (Newton's Second Law Of Motion)

So, if F= O, then it doesn't matter what mass is.

0= ma

The projectile has mass, but Acceleration will be zero because anything multiplied by zero is zero

m= O * accn

So, there is no acceleration, because there are no forces(9 votes)

- There is said:"we figured out... in the last video..." in this video. My question is: In what last video. There were another conditions in previous video "Launching and landing on different elevations".(6 votes)
- You can find the first part of the video on youtube:

https://www.youtube.com/watch?v=UXORCpvKYiI&index=6&list=PLKQ6Y3lMA4BZtUVofapAlfd7vLsbiFpTa(6 votes)

- Why does Sal use the initial component of the velocity of 29.54m/s when it seems it should 0m/s from attaining its highest point? I thought the initial velocity would be from the apex of its trajectory since that is the immediate initial velocity before the projectile lands..(5 votes)
- We know the time that passed since the launch of the projectile, not the time that passed since it's highest point. So we're using the total change of time, from the very beginning, that's why we also use the launch velocity. You could calculate it from the apex, but then you have to first figure out the change of time from the apex to landing, which will be less than 5.67 s(6 votes)

- Two questions:

1. We will always take the final velocity vertically when calculating the total velocity?

2. Why did Sal not take the -26.03 value instead of +26.03??(7 votes)- 1) Yes, even if it 0m/s. Although, if it were, we wouldn't really need to take it anyways.

2) 1-dimensional planes have only 2 directions (left, right, etc.) which are described using + and - symbols.

For 2 dimensions, it becomes complicated because there are multiple directions. Using + and - symbols would not work. So, directions are described using angles from the horizontal line.

But, regardless of this idea, or maybe because of it, Pythagoras' Theorem is used. And from the start, this theorem is only suited to find the magnitude of the vector. It is not suited to include directions in the calculation.

(Adding a negative sign presumably gives a false answer. Because in real life, even if we point the horizontal line in the opposite direction, the hypotenuse's length would be the same)

And finally, when the magnitude is obtained, it can be used alongside the angle/direction to describe the vector.(2 votes)

- At3:31Sal calculates -26.026. When I subtract 9.8 from 29.54 and multiply that by 5.67, I get 111.93. Why do I not get -26.026? Is my math incorrect?(2 votes)
- For calculating the final vertical velocity, is it possible to use the formula: displacement=((initial velocity + final velocity)/2)*change in time? After substituting all the known values in, I get the result −26.017 which is only slightly different from Sal's result. Could this be because of small rounding errors or is this not the right equation to use?(2 votes)
- the questions to ask your self are:

Is the acceleration constant in this situation? does my equation include all the known and just one unknown variables.

Seems to me you got it right.

What you think??(2 votes)

- How would you do this if you were accounting for air resistance ?(1 vote)
- What is the horizontal and why is it the angle below the horizontal? The object lands higher than where it was launched, so I'm confused about why the direction of the velocity would be "below" or negative.(1 vote)
- We are talking about the velocity here, not the displacement. The object first went up and, when it hit the top platform, it's on its way down again. Therefore the direction of its movement over time (definition of velocity) is downward, with that angle. Hope this helps.(3 votes)

## Video transcript

In the last video, I told
you that we would figure out the final velocity of
when this thing lands. So let's do that. I forgot to do it
in the last video. So let's figure out
the final velocity-- the vertical and the
horizontal components of that final velocity. And then we can reconstruct
the total final velocity. So the horizontal
component is easy, because we already know that
the horizontal component of its velocity is this value
right over here, which we-- this 30 cosine of 80 degrees. And that's not going to
change at any point in time. So this is going to be
the horizontal component of the projectile's
velocity when it lands. But what we need
to do is figure out the vertical component
of its velocity. Well, one thing we did
figure out in the last video, we figured out what the time
in the air is going to be. And we know a way of figuring
out our final velocity from an initial velocity
given our time in the air. We know that a
change in velocity -- and we're only dealing
in the vertical now-- we're only dealing
with the vertical, because the horizontal velocity
is not going to change. We've assumed that air
resistance is negligible. So we're only dealing with
the vertical component right over here. We know that the
change in velocity-- or, we could say
the horizontal-- the vertical component of
the change in velocity, is equal to the vertical
component of the acceleration times time. Now, we know what
the change in time is, we know it is-- I'll just
write down times our time. And what is our
change in velocity? Well, our change in velocity
is our final vertical velocity minus our initial
vertical velocity. And we know what our
initial vertical velocity is, we solved for it. Our initial vertical
velocity, we figured out, was 29.54 meters per second. That's 30 sine of 80 degrees,
29.54 meters per second. So this is going to be minus
29.54 meters per second, is equal to-- our acceleration
in the vertical direction is negative, because
it's accelerating us downwards, negative 9.8
meters per second squared. And our time in the
air is 5.67 seconds. Times 5.67 seconds. And so we can solve for
the vertical component of our final velocity. So once again, this is
the vertical component. This isn't the total one. So, the vertical component. Let me-- well I wrote
vertical up here. So there's the
vertical component. So let's solve for this. So if you add 29.54
to both sides, you get the vertical component
of your final velocity. Well, this is a
vertical component, I didn't mark it
up here properly-- is equal to 29.54 meters
per second plus 9.8 plus -- or I should say minus--
meters per second. Minus 9.8 meters per second
squared, times 5.67 seconds. The seconds cancel out
with one of these seconds. So everything is
meters per second. And so, get the
calculator out again, we have 29.54 minus
9.8 times 5.67. So we get our change-- our final
velocity is negative 26.03. So this is negative
26.03 meters per second. And you might say
wait, wait Sal , what is this negative 26.03
meters per second mean? Remember, when we're dealing
in the vertical dimension, positive means up,
negative mean down. So it means we're going 26.03
meters per second downwards. Downwards, right when we land. So what is our
total velocity when we fall back to that landing? So the vertical
component of our velocity is negative 29.06 times .03
in the downward direction. And the horizontal component
of our velocity, we know, hadn't changed the entire time. That, we figured out, was
30 cosine of 80 degrees. So that over here, is
30 cosine of 80 degrees. I'll get the calculator
out to calculate it. 30 cosine of 80 degrees,
which is equal to 5.21. So this is 5.21
meters per second. These are both in
meters per second. So what is the total velocity? Well, I can do
the head to tails. So I can shift this guy
over so that its tail is at the head of the blue vector. So it would look like that. The length of
this-- the magnitude of our vertical
component, is 29.03. And then we could just use
the Pythagorean theorem to figure out the magnitude of
the total velocity upon impact. So the length of
that-- we could just use the Pythagorean theorem. So the magnitude of our
total velocity, that's this length right over here. The magnitude of
our total velocity, our total final velocity
I guess we can say, is going to be equal to-- well
that's-- let me write it this way. The magnitude of
our total velocity is going to be equal to square
root-- this is just straight from the Pythagorean
theorem-- of 5.21 squared plus 29.03 squared. And we get it as
being the second-- the square root of 5.21
squared plus 29.03 squared gives us 29.49
meters per second. This is equal to 29.49
meters per second. That is the magnitude
of our final velocity, but we also need to
figure out its direction. And so we need to
figure out this angle. And now we're talking about
an angle below the horizontal. Or, if you wanted to view
it in kind of pure terms, it would be a
negative angle-- or we could say an angle
below the horizontal. So what is this angle
right over here? So if we view it as
a positive angle just in the traditional
trigonometric way, we could say that the-- we could
use any of the trig functions, we could even use tangent. Let's use tangent. We could say that the
tangent of the angle, is equal to the opposite
over the adjacent-- is equal to 29.03 over 5.21. Or that theta is equal
to the inverse tangent, or the arctangent
of 29.03 over 5.21. And that gives us-- we take
the inverse tangent of 29.03 divided by 5.21, and
we get 79.8 degrees. But it's going to be
79.8 degrees south, or, I guess, below
the horizontal. Or you could view this as an
angle of negative 79.8 degrees above the horizontal,
either one of those work. What's neat about this, is we
figured out our final velocity vector. The entire vector, we know
what that entire vector is. It is 29.49 meters
per second at 79.8 degrees below the horizontal.