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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1 > Unit 1

Lesson 7: Objects in freefall# Plotting projectile displacement, acceleration, and velocity

Plotting projectile displacement, acceleration, and velocity as a function of time. Created by Sal Khan.

## Want to join the conversation?

- If acceleration is always constant, when the ball is in its peak, would the acceleration be zero?(21 votes)
- The velocity is zero at the peak, but not the acceleration. If the acceleration were also zero, the ball would not fall back down and stay in the air. The acceleration from gravity causes the velocity to change in the downward direction.(68 votes)

- I didn't understand how a ball thrown in the air could have an initial velocity. Can you please give an example?(17 votes)
- Initial velocity in this case refers to the velocity of the ball with which it is thrown in the air .A ball thrown in the air not having initial velocity would mean holding the ball in the hand and then removing our hand just leaving it to the mercy of the earth's gravity. In order for the ball to move up a force has to be applied which is against the gravitational pull and the applied force results in velocity to the ball.(35 votes)

- It may be a silly question but can somebody help me here? When the ball hits the ground, I just can't figure out how come the velocity of the ball is not zero (even for a short while - same as velocity of ball at its peak in air). I understand it would hit the ground hard but couldn't it have to stop shortly before bouncing back? Thanks(7 votes)
- No question is a silly question, Pham. So, the final velocity is the velocity of the object right before it touches the ground. It is not the velocity after hits the ground.(11 votes)

- In the displacement formula shown in the video (S = Vi*t+1/2a(t)^2), Kahn only applies the 1/2 to the acceleration, not time squared. In other words, when time is 2 seconds, he takes half of the -9.8m/s, but then uses the full 4 seconds. Is that correct, or does the 1/2 not get distributed to the full value (including time)?(5 votes)
- Its actually 1/2*a*t*t (=(at^2)/2) so its the same thing.

So when you simplify 1/2*9.8*t*t you get -4.9*t*t

hope it made sense(10 votes)

- if a ball is thrown upwards from the top of a tower and then falls on the ground ,is acceleration g or -g(7 votes)
- You can always define direction however you want, as long as you are consistent throughout your problem. If you define down as positive, then acceleration is 9.8 m/s^2. If you define up as positive, then acceleration is -9.8 m/s^2.(7 votes)

- Are we missing a video? I’m kinda confused 😂(8 votes)
- I agree, it feels like there was supposed to be a video before, but it wasn't included(3 votes)

- I learned that acceleration due to gravity is 9.81 m/s^2. How much of a difference would it from his answers using just 9.8?(2 votes)
- Gravity acceleration on Earth's surface varies anywhere from 9.78 to 9.83 m/s^2 depending on where you are, so it doesn't make much sense to go beyond 9.8 m/s^2.(5 votes)

- i don't understand why didn't Sal calculate the mas of the object? wouldn't it mean that if we drop 2 objects, the object with a greater mass and the object with the smaller mas will fall at the same time if they fell at the same time and same distance?(1 vote)
- On earth, the mass of the object does actually make a difference to the rate of acceleration. This is due to air resistance. However, in mathematics, calculations are usually made without factoring in air resistance. Scientists once conducted an experiment on the moon (there is no air on the moon) where an astronaut dropped a feather and a hammer, from the same height and at the same time. Both the feather and the hammer landed on the ground at exactly the same time.(9 votes)

- Can the acceleration be considered as the derivative of the velocity? Can the velocity be considered as the derivative of the displacement?(4 votes)
- So, a couple of videos ago, Sal taught us that with constant acceleration over a specified period of time on a V/T graph (Velocity as Y-axis, Time as X-axis), the data will form a straight-line slope; this is because acceleration was/is constant for that video. The distance travelled is equal to the area beneath the slope (should form a right triangle.) My question is: are there any rules, such as that one which apply in this situation, given a parabolic shape, versus a right triangle?(0 votes)
- Distance is always the area under the velocity graph, regardless of the shape of the graph. It just gets more challenging to calculate what that area is. That's where integral calculus comes in.(6 votes)

## Video transcript

What I want to do in this video,
now that we have displacement as a function of time,
given constant acceleration and an initial velocity. I want to plot displacement,
final velocity, and acceleration, all of
those as functions of time. Just so we really
understand what's happening as the ball is
going up and then down. So we know this is
our displacement as a function of time. We know what our final
velocity is going to be, as a function of time. We talked about it
in the last video. Our final velocity is going
to be our initial velocity plus our acceleration
times change in time. Right? If we start at some initial
velocity, and then you multiply the
acceleration times time. This part tells you how
much faster or slower you're going to go than
your initial velocity. And that will be, I
guess you could say, your current velocity,
or the final velocity at that point in time. And of course, our
acceleration, we know. Our acceleration is
pretty straightforward. The acceleration due
to gravity is just going to be negative 9.8
meters per second squared. Once again, negative
being the convention that it is in the
downward direction. Our initial velocity
is going to be in the upward direction,
19.6 meters per second. So let's plot these
out a little bit. Let's plot these out. So the first graph I want
to do, right over here, will be my displacement
versus time. So this axis right over
here is going to be time, or maybe I could call this
the change in time axis. Actually, lets
just call it time. And then this axis, right over
here, I will call displacement. And let me put
some markers here. So let's say that this is
5 meters, 10 meters, 15 meters, and 20 meters. And then in the time, this is 0,
this is 1 this is 2, this is 3, and this is 4 seconds. So this is in
seconds right here. This is meters. 5, 10, 5, 10, 15, 20. So this is displacement. Displacement graph. And I want to, at the
same time as that, I want to do a velocity graph. So let me draw my
velocity graph like this. I'll do it a little
bit different. So, this is because the velocity
will be going up and down. So we need to have positive
and negative values here. Time will only be positive. So once again, I care about 1
second, 2 seconds, 3 seconds, and 4 seconds in time. And velocity, I'm
going to call this. This is going to be
10 meters per second. This is 20 meters per second. This will be negative
10 meters per second. And this will be negative
20 meters per second. And so all of this is
in meters per second. This right here is velocity. This axis right here is time. So this is my velocity graph. And why don't we just throw an
acceleration graph over here, although that's, to some
degree, the easiest of them all. So the acceleration
graph, and I'll just do this right
from the get go, because we're going to assume
that acceleration is constant. So this is 1 second,
2 seconds, 3 seconds, and 4 seconds into it. And then let's call
this negative 10. And all of this is in
meters per second squared. And so we know our
acceleration is negative 9.8 meters per second squared. So the acceleration
the entire time over the four seconds, the
acceleration over the four seconds is going to be
that's about negative 9.8. It's going to be that. It's going to be a constant
acceleration the entire time. But let's figure out
displacement and velocity. So let me draw a
little table here. So in one column, I
will do change in time. Or sometimes you
could do that as time. Let's figure out what
our final velocity is, or I should really say our
current velocity, or velocity at that time. And then in this
column, I'll figure out what our displacement is. And I will do it for
times 0, 1, 2, 3, 4. Or change in time. So when 0 seconds have gone
by, when 1 second has gone by, when 2 seconds, 3 seconds,
and 4 seconds have gone by. Actually let, me call this
the change in time axis, because this is essentially
how many seconds have gone by. So this is my
change in time axis. And let me make it clear that
this graph-- I didn't label it here-- this is my
acceleration graph. And I'm going off of the screen. All right. So let's fill these things out. So at time 0, what is
our what is our velocity? Well, if we use this expression
right here, time 0, or delta t is equal to 0. This expression right
here, is going to be 0. And it's just going to
be our initial velocity. And in the last video, we
gave our initial velocity, is going to be as 19.6
meters per second. So it is going to be
19.6 meters per second. And let me plot that over here. At time 0, it is going to
be 19.6 meters per second. What is our initial displacement
at time zero, or change in time 0? So you look at this up here. Well, our delta t is 0, so this
expression is going to be 0, and this expression
is going to be 0. So we haven't done
any displacement yet, when no time has gone by. So we have done no displacement. We are right over there. Now, what happens after
1 second has gone by? What is now our velocity? Well, our initial velocity,
right over here, is 19.6. 19.6 meters per second,
that was a given. And our acceleration is negative
9.8 meters per second squared. So it's a negative,
right over there. And then you multiply that times
delta t in every situation. So in this situation
we're going to multiply it by 1, because delta t is 1. So we get 19.6 minus
9.8, that gives us exactly 9.8 meters per second. And the units work out,
because you multiply this times seconds, this gives
you meters per second. So 19.6 meters per second minus
9.8 meters per second-- one of these seconds goes away when
you multiply it by second-- gives you 9.8 meters per second. So after 1 second,
our velocity is now half of what it was before. So we're now going
9.8 meters per second. Let me draw a line here. 9.8 meters per second. Now what is our displacement? So you look up here. And let me rewrite this
displacement formula, with all of the
information that we know. So we know that
displacement is going to be equal to our initial
velocity, which is 19.6-- and I won't write
the units here, just for the sake of space--
times our change in time. I'll do it in that same
color, so you what's what. Times our change
in time plus 1/2. Now let me be clear, 1/2
times negative 9.8 meters per second squared. So 1/2 times a is
going to be-- actually I can rewrite this,
right over here-- because this is going to
be negative 9.8 meters per second times 1/2, so this
is going to be negative 4.9. All I did, is I took 1/2
times negative 9.8 over here. 1/2 times negative 9.8. And this is important. And this is why the vector
quantities start to matter. Because if you didn't, if
you put a positive here, you wouldn't have
the object actually slowing down as it went up,
because you would have gravity somehow accelerating
it as it went up. But it's actually
slowing it down. It's accelerating it in
the downwards direction. So that's why you have to have
that negative right over there. That was our convention at the
beginning of the last video. Up is positive. Down is negative. So let's focus. So this part right over
here, negative 4.9 meters per second squared
times delta t squared. And this will make it
a little bit easier, although we'll still-- let
me get the calculator out. So when one second has passed--
I'll get my trusty TI-85 out. When one second has passed, the
displacement is 19.6 times 1. Well, that's just 19.6. Minus 4.9 times 1 squared. So that's just minus 4.9. Gives us 14.7 meters. So after 1 second, the ball
has traveled 14.7 meters in the air. So that's roughly over there. Now, what happens
after 2 seconds? I'll do this in magenta. So after 2 seconds, our velocity
is 19.6 minus 9.8 times 2. 2 seconds have gone by. Well, 9.8 meters per second
squared times 2 seconds gives us 19.6 meters per second. So these just cancel out. So we get, our
velocity is now 0. So after 2 seconds,
our velocity is now 0. Actually, let me-- So let me
make it so it's-- this thing should look more like a line. I don't make you get a sense--
So this is-- So let me just draw the line like this. Our velocity is now
0 after 2 seconds. What is our displacement? So we're literally at the point
where the ball has no velocity. At exactly 2 seconds. So it's kind of gone up, and
for that exact moment in time, it is stationary. And then what do we have
going on in our displacement? We have 19.6-- let me get
the calculator out for this. I could do it by
hand, but for the sake of quickness-- 19.6
times 2-- 19.6 times 2 seconds minus 4.9
times 2 seconds squared. This is 2 seconds squared. Oh, I lost the calculator. Times 2 seconds squared. So that's times 4. So that gives us 19.6 meters. So we're at 19-- let me
do that in magenta-- we are at 19.6 meters. So after 2 seconds, we are
19.6 meters in the air. Now let's go to 3 seconds. So after 3 seconds,
our velocity is now-- I'll just get the-- it's
19.6 meters per second minus 9.8 times 3. And we could do
that in our head, but just to verify it for us,
let me get the calculator out. It's 19.6 minus 9.8 times 3. That gives us negative
9.8 meters per second. So after 3 seconds,
our velocity is now negative 9.8 meters per second. What does that mean? It's now going in the
downward direction at 9.8 meters per second. So this is our velocity graph. And then what is our
displacement at this point? So once again, let's
get the calculator out. If you're getting
the hang of this. At any time, I encourage
you to pause it, and try it for yourself. So now, what is-- now
this is a little, OK. So I'm looking at
my displacement, I wrote right over here. So our displacement
where delta t is 3 seconds, 19.6 times 3
minus 4.9 times-- and this is delta t, so
this is 3 seconds. We're talking about when delta
t, or our change in time, is 3 seconds. So that squared. So times 9. And that gives us 14.7 meters. So after 3 seconds, we're
at 14.7 meters again. And so we're at the same
position we were at 1 second, but the difference is, now
we're moving downwards. Over here we were
moving upwards. And then finally, what
happens after 4 seconds? Well, what's our velocity? Well let me just get
the calculator out, although you might be able to
figure this out in your head. Our velocity is going to be
19.6 minus 9.8 times 4 seconds. Which is minus 19.6
meters per second. So our magnitude of our
velocity is the same as when we initially threw
the ball, except now it's going in the opposite direction. It's now going downwards. And what is our displacement? Get the calculator out. So we have, our displacement
is 19.6 times 4-- 4 seconds have gone by-- minus
4.9 times 4 squared-- which is 16-- so times
16, which is equal to 0. Our displacement is 0. We are back on the ground. We are back on the ground. So if you were to
plot its displacement you would actually get a
parabola, a downward opening parabola that looks
something like this. Doing my best to draw
it relatively neatly. Actually, I can do a
better job than that. I'll do a dotted line. Dotted lines are always
easier to adjust midstream. If you plot displacement
versus time, it looks something like this. Its velocity is just this
downward sloping line, and then the
acceleration is constant. And the whole reason
why I wanted to do this, is I wanted to show you that
the velocity the whole time is decreasing at
a constant pace. And that makes sense, because
the rate at which the velocity increases or decreases
is the acceleration. And the acceleration, based on
our convention, is downwards. So that's why it's decreasing. We have a negative slope here. We have a negative slope
of negative 9.8 meters per second squared. And so just to think
about what's happening for this ball, or
this rock-- and I know this video is
getting long-- as it goes through the air, I'm going to
draw the vectors for velocity. And I'm going to
do that in orange, or no, maybe I'll
do that in blue. So velocity in blue. So right when we start,
it has a positive velocity of 19.6 meters per second. So I'll draw a big
vector like this, 19.6 meters per second,
that's its velocity. Then after 1 second, it's
9.8 meters per second. So it's half of that. Maybe it would look
something like this, 9.8 meters per second. Then at this peak right over
here, it has a velocity of 0. Then as you go to 3 seconds,
the magnitude of its velocity is 9.8 meters per second. But it is now downwards. It is now downwards,
so it looks like this. And then, finally, right when
it hits the ground, right before it hits the ground,
it has a negative velocity of 19.6 meters per second. So it would look like
this, roughly like this. If I use the same
scale over here. But what was the
acceleration the entire time? Well, the entire time, the
acceleration was negative. It was negative 9.8
meters per second squared. And I'll do that in orange. So the acceleration,
over here, negative-- no, I want to do that in
orange-- the acceleration was negative 9.8 meters
per second squared. Acceleration, negative 9.8
meters per second squared. Negative 9.8 meters
per second squared. The acceleration is
constant the entire time. This last one is negative 9.8
meters per second squared. It does not change,
depending where you are in the
curve, when you're near the surface of the Earth. So hopefully that clarifies
things a little bit and gives you a
good sense of what happens when you throw a
projectile into the air.