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## Kinematic formulas and projectile motion

Current time:0:00Total duration:7:09

# Deriving max projectile displacement given time

## Video transcript

Just want to follow up
on the last video, where we threw balls in the air, and
saw how long they stayed up in the air. And we used that to figure
out how fast we initially threw the ball and how
high they went in the air. And in the last video, we
did it with specific numbers. In this video, I
just want to see if we can derive some
interesting formulas so that we can do the computations
really fast in our brains, while we're playing this game
out on some type of a field, and we don't necessarily
have any paper around. So let's say that the ball
is in the air for delta t. Delta t is equal
to time in the air. Then we know that
the time up is going to be half that, which is the
same thing as the time down. The time up is going to
be equal to delta t-- I'm going to do that
in the same color-- is going to be equal to the
time in the air divided by 2. So what was our
initial velocity? Well, all we have to
do is remind ourselves that the change
in velocity, which is the same thing as
the final velocity minus the initial velocity. So the final
velocity-- remember, we're just talking about half
of the path of this ball. So the time that
it gets released, and it's going at kind of
its maximum upward velocity, and it goes slower, and slower,
slower, all the way until it's stationary for just
a moment, and then it starts going down again. Now remember, the acceleration
is constant downwards this entire time. So what is the final
velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0
minus our initial velocity, when it was taking off. That's our change in velocity. This is our change
in velocity, is going to be equal to the
acceleration of gravity, negative 9.8 meters
per second squared-- or the acceleration
due to gravity when an object is in free fall,
to be technically correct-- times the time that
we are going up. So times delta t up,
which is the same thing. I won't even write it. Delta t up is the same
thing as our total time in the air divided by 2. And so we get negative
the initial velocity is equal to-- this thing,
when you divide it by 2, is going to be 4.9 meters
per second squared-- we still have our negative
out front-- times our delta t. Remember, this is our
total time in the air, not just the time up. This is our total
time in the air. And then we multiply both
sides times a negative. We get our initial
velocity is just going to be equal to 4.9
meters per second squared times the total time that
we are in the air. Or you could say
it's going to be 9.8 meters per second squared
times half of the time that we're in the air. Either of those would get
you the same calculation. So let's figure out our total
distance, or the distance that we travel in the time up. So that'll give us
our peak distance. Remember that distance-- or
I should say displacement, in this situation--
displacement is equal to average velocity
times change in time. The change in time that we
care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average
velocity, if we assume constant acceleration,
is your initial velocity plus your final velocity over 2. It's really just the
mean of the two things. Well, we know what our
initial velocity is. Our initial velocity is
this thing over here. So this is this thing over here. Our final velocity--
remember, we're just talking about
the first half of the time the
ball is in the air-- so it's final velocity is 0. We're talking when it gets
to this peak point, right over here-- that's from two
videos ago-- that peak point right over there. So our average
velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters
per second squared times delta t over 2. So this right here, this
is our average velocity. Velocity average. So let's stick that
back over here. So our maximum
displacement is going to be our average
velocity-- so that is 4.9 meters per second
squared-- times delta t, all of that over 2. And then we multiply it
again times the time up. So times delta t over 2 again. This is the same thing. These are the same thing. And then we can simplify it. Our maximum displacement is
equal to 4.9 meters per second squared times delta t
squared, all of that over 4. And then we can just
divide 4.9 divided by 4. 4.9 divided by 4 is-- let me
just get the calculator out. I don't want to do that
in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum
displacement is going to be 1.225 times our total
time in the air squared, which is a pretty
straightforward calculation. So this is our max
displacement, kind of how high we get displaced. Right when that
ball is stationary, or has no net velocity, just
for a moment, and starts decelerating downwards. So we can use that. If a ball is in the air for
5 seconds-- we can verify our computation from
the last video-- our maximum displacement, 1.225,
times 5 squared, which is 25, will give us 30.625. That's what we got
in the last video. If the ball's in the
air for, I don't know, 2.3 seconds-- so
it's 1.225 times 2.3 squared-- then that means it
went 6.48 meters in the air. So anyway, I just wanted to give
you a simple expression that gives you the maximum
displacement from the ground, assuming air resistance
is negligible, as a function of the
total time in the air. I don't know. I find that pretty fun. And it's a neat game to play.