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### Course: Physics library > Unit 1

Lesson 4: Kinematic formulas and projectile motion- Average velocity for constant acceleration
- Acceleration of aircraft carrier take-off
- Airbus A380 take-off distance
- Deriving displacement as a function of time, acceleration, and initial velocity
- Plotting projectile displacement, acceleration, and velocity
- Projectile height given time
- Deriving max projectile displacement given time
- Impact velocity from given height
- Viewing g as the value of Earth's gravitational field near the surface
- What are the kinematic formulas?
- Choosing kinematic equations
- Setting up problems with constant acceleration
- Kinematic formulas in one-dimension

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# Projectile height given time

Figuring out how high a ball gets given how long it spends in the air. Created by Sal Khan.

## Want to join the conversation?

- so does this mean that if I project an elephant with a velocity of 24.5 m/s, it is going to reach the same height as a feather projected with the same velocity ?(132 votes)
- without air, yes.

but on earth air and shape of feather does not allow you to do that

also to project an elephant with that velocity would be such a pain.(355 votes)

- why is velocity final zero??(29 votes)
- when you threw a ball up watch and it kind of stops in the air that's the peak(12 votes)

- I don't see how it's possible to derive the initial velocity on the up trip from knowing the velocity on the down trip. If I shoot a bullet in the air, it's going to travel faster on the up trip than on the down trip, isn't it? On the up trip it may go 24.5m/s^2 in 1s and on the down trip 9.8m/s^2 in 2.5s. Same overall displacement, but different velocities and accelerations. Unless I'm missing something...(8 votes)
- The total distance that bullet travels vertical is equal in this case to the total distance travelled up and down. The inital velocity is going to be 'slowed' down to zero (m/s) because of gravity, and the effect will be equally the same returning. Therefore the acceleration/deaccelertion is equl, the distance is equal and so will the time. This will ultimately show that the velocity at the exact same elevation (height) will be the same.

If, the bullet hits and object higher than it was orginally fired, it will be travelling that much slower, and vice versa if the bullet travels below (say of a cliff) it will continue accelerating until impact.(7 votes)

- Could anyone please confirm the displacement of 30.6 meters is only for half the time. i.e. distance reached at halfway point. or is it the total distance?(3 votes)
- JULIAN if SAL sir considered 5 sec that if ball came down then distance would be 30.2*2 but displacement would be 0m as it returned from where it started and you can see that Sal has taken only 2.5s which means distance traveled during its journey to apex height(9 votes)

- So what exactly is air resistance, and why does everyone treat it as "negligible"?(3 votes)
- There are two main effects that are part of air resistance.

One is the need for the object moving through the air to move the air out of the way as it move through it.

The second effect is drag which is caused as the air flows around an object there is usually a low pressure region created behind the object that acts to pull back on the object. The size of this effect depends on the overall shape of the object, there is less drag on a teardrop shaped object than a box.

Air resistance is treated as negligible in many problems because air has a low density and at speeds that are a small fraction of the speed of sound air resistance is a small effect.(6 votes)

- At3:40is the initial velocity zero or the final velocity? Earlier in the video he said that the initial velocity is zero.(4 votes)
- Even I was stuck at this but later got it:)

Sal said that the initial velocity is zero because the ball wasn't thrown yet and it was at**rest**.

Later he said that the final velocity is zero. Huh? Because when the ball was thrown up and it reached its**final position**, it again had to come down . Its new initial velocity was zero again because it had to come all the way down*again*and that was the`final velocity`

.

Hope that helps:)(3 votes)

- When you are calculating displacement at the end... where do you get the equation to solve for it. I used the equation v(f)^2 = v(Initial)^2 + 2a(change in x)... However I get 3.125 meters instead... which should get the same value for the change in x as you but I didn't. What would I have to do differently to get the displacement of the object?(4 votes)
- Well I did the calculations and for both V(F)^2 = V(0)^2 + 2ad and d = V(F) + V(0)/2 * t

Here's what I got:

d = V(F) + V(0)/2 * t

d = (-24.5 + 0)/2 * 2.5

d = -12.25 * 2.5

d = -30.625

V(F)^2 = V(0)^2 + 2ad

-24.5^2 = 0^2 + 2(-9.8)d

600.25 = (-19.6)d

d = 600.25/-19.6

d = -30.625

The ball's initial velocity is the one measured at its highest point and its final velocity is the one right before it hit the ground, going downwards. In this case, the up is positive so having a negative displacement would mean it went down from its initial position.(1 vote)

- is the velocity calculated by Average on6:25because both the rise and fall of the object is taken into question?(2 votes)
- No, Van. We are only dealing with the first part of the projectile-rising part, in which t=2.5s. This is because the falling part is essentially the reverse of rising part. We use average velocity because velocity is not constant, but acceleration is.(4 votes)

- in the last video sal said that displacement is equal to initial velocity times time + 1/2 acceleration times time squared, but in this video an else formula was used by him, why?(3 votes)
- I noticed in the first video the final velocity was unknown. This video the final velocity is 0. I think so anyway(1 vote)

- Is air resistance only negligible because it complicates the answer and equations?(2 votes)
- Yes, and in some real-life situations air resistance is small enough that we don't need to worry about it. For example if you through a heavy metal ball up in the air, you would probably not be able to measure the difference between doing it in a vacuum or in air. That's because the surface area of the ball is small compared to its mass, and it has a smooth shape, and the velocity is not very high if you are just throwing it with your hand. In some cases, the difference might be measurable but still too small for you to want to worry about.

It is very complex to account for air resistance precisely, and that would really interfere with trying to learn the basic concepts and math involved with how objects move. You have to learn to walk before you can learn to run.(2 votes)

## Video transcript

Let's say you and I are
playing a game where we're trying to figure out how high a
ball is being thrown in the air or how fast that we're
throwing that ball in the air. And what we do is
one of us has a ball and the other one has
a stopwatch over here. So this is my best attempt. It looks more like a
cat than a stopwatch, but I think you get the idea. And what we do is one
of us throws the ball, and the other one times how
long the ball is in the air. And then what we
do is we're going to use that time in
the air to figure out how fast the ball was
thrown straight up, and how long it was in the
air and how high it got. And there's going to be one
assumption I make here-- and frankly, this
is an assumption that we're going to make
in all of these projectile motion-type problems-- is that
air resistance is negligible. And for something like-- if
this is a baseball, or something like that, that's a
pretty good approximation. So we're not going to
get the exact answer. And I encourage you to
experiment on your own to see what air resistance does
relative to your calculations. But we're going to assume
for this projectile motion-- and really all of
the future ones or at least in the
basic physics playlist-- we're going to assume that
air resistance is negligible. And what that does for
us is that we can assume that the time for the ball
to go up to its peak height is the same thing as the time
that it takes to go down. If you look at
this previous video where we plotted
displacement versus time, you see after 2 seconds the ball
went from being on the ground-- or I guess the thrower's hand--
all the way to its peak height. And in the next 2 seconds, it
took that same amount of time to go back down to the
ground, which makes sense. Whatever the
initial velocity is, it takes half the
time to go to 0. And it takes that
same amount of time to now be accelerating
in the downward direction back to that same
magnitude of velocity but now in the
downward direction. So let's play around
with some numbers here just so we get a little
bit more of a concrete sense. So let's say I throw
a ball in the air. And you measure,
using the stopwatch, that the ball is in
the air for 5 seconds. So how do we figure out
how fast I threw the ball? Well, the first
thing we can do is we could say, look, if
the total time in the air was 5 seconds, that means that
the time-- let me write it. That means that
the change in time to go up during the
first half of, I guess, the ball's time in the air
is going to be 2.5 seconds. Which tells us that
over this 2.5 seconds, we went from our initial
velocity, whatever it was, we went from our
initial velocity to our final velocity,
which is a velocity of 0 meters per second
in the 2 and 1/2 seconds. And this isn't the
graph for that example. This is the graph for the
previous example where we knew the initial velocity. But in whatever
that time is, you're going from your initial velocity
to being stationary at the top, right? Right when the ball
is stationary then it starts getting
increasing velocity in the downward direction. So it takes 2.5 seconds to
go from some initial velocity to 0 seconds. So we do know what the
acceleration of gravity is. We know that the acceleration
of gravity here-- we're assuming it's constant,
although it's slightly not constant. But we're going to
assume it's constant, if we're just dealing close
to the surface of the earth-- is negative 9.8 meters
per second squared. So let's think about it. Our change in velocity
is the final velocity minus the initial
velocity, which is the same thing as 0 minus
the initial velocity, which is the negative of
the initial velocity. And what's another way to
think about change in velocity? Well, just from the
definition of acceleration, change in velocity is
equal to acceleration-- negative 9.8 meters per
second squared-- times time, or times change in time. We're just talking
about the first half of the ball's time in the air. So our change in time is 2.5
seconds-- times 2.5 seconds. So what is our
change in velocity, which is also the same
thing as the negative of our initial velocity? Let me get my calculator,
bring it onto the screen. So it is negative 9.8 meters
per second times 2.5 seconds. It gives us negative 24.5. So this gives us-- let me
write it in a new color. This gives us negative
24.5 meters per second. This second cancels out
with one of these seconds in the denominator, so we only
have one in the denominator now, so it's meters per second. And this is the same thing as
the negative initial velocity. That's the same thing as
our change in velocity. And so you multiply both
sides by a negative. We get our initial velocity. So that simply we were
able to figure out what our velocity was. So literally you take the
total time in the air, take it and divide
it by 2, and then multiply that by the
acceleration of gravity. And I guess you could take
the absolute value of that or you take the positive
version of that. And then that gives you
your initial velocity. So your initial velocity
here is literally 24.5 meters per second. And since it's a
positive quantity, it is upwards in this example. So that's my initial velocity. So we already figured
out part of this game, the initial velocity
that I threw it upwards. And that's also
going to be-- we're going to have the same
magnitude of velocity when the ball's about
to hit the ground, although it's going to be
in the other direction. So what is the distance--
or let me make it clear. What is the displacement of the
ball from its lowest point-- right when it leaves your
hand-- all the way to the peak? Well, we just have to remember--
and once again, all this comes from very straightforward
ideas, change in velocity is equal to acceleration
times change in time. And then the other
simple idea is that displacement is equal
to average velocity times change in time. Now, what is our
average velocity? Our average velocity is
your initial velocity plus your final
velocity divided by 2, if we assume
acceleration is constant. So it's literally just
the arithmetic mean of your initial and
final velocities. So what is that? That's going to be 24.5
meters per second plus-- what's our final velocity? In this situation,
remember, we're just going over the
first 2.5 seconds. So our final velocity is once
again 0 meters per second. We're just talking
about when we get to this point right over here. So our final velocity is
just 0 meters per second. And we're just going
to divide that by 2. This will give us
the average velocity. And then we want to multiply
that times 2.5 seconds. So we get-- this part right
over here-- 24.5 divided by 2. We can ignore the 0. That still is 24.5 That
gives us 12.25 times 2.5. And remember, this right
over here is in seconds. Let me write the units down. So this is 12.25 meters per
second times 2.5 seconds. And just to remind
ourselves, we're calculating the displacement
over the first 2 and 1/2 seconds. So this gives us-- I'll get
the calculator out once again. We have 12.25 times 2.5
seconds gives us 30.625. So this gives us--
so our displacement is 30.625 meters-- these
seconds cancel out-- meters. This is actually a ton. This is roughly, give or
take, about 90 feet thrown in the air. So this would be like
a nine-story building. And I, frankly, do not
have the arm for that. But if someone is
able to throw the ball for 5 seconds in
the air, they have thrown it 30 meters in the air. Well, hopefully you
found that entertaining. In the next video,
I'll generalize this. Maybe we can get a
little bit of a formula so maybe you can generalize it. So regardless of the
measurement of time, you can get the
displacement in the air. Or even better, try to
derive it yourself and we'll see how at least I tackle
it in the next video.