Mr. P: Good Morning.
Today we are going to begin an introductory
projectile motion problem where the object has
an initial velocity that is entirely horizontal,
and we are solving for the displacement in the x direction.
(dramatic music) Here we go.
Bo: Hey guys.
Billy: Hey Bo.
Bobby: Hi Bo.
♫ (lyrics) Flipping physics ♫
Mr. P: Bo, could you please read the problem?
Bo: Sure Mr. P, I can read the problem.
While in a car moving at 10.0 miles per hour,
Mr. P drops a ball from a height of 0.70 meters
above the top of a bucket.
Part One, how far in front of the bucket
should he drop the ball such that the ball
will land in the bucket?
Part One?
Where are the other parts?
Mr. P: Bo, after we complete this part
we'll talk about the next one.
(dramatic music)
Billy: That is Mr. P holding the ball, is it?
Bobby: I think it's one of his kids.
(marker squeaking)
Mr. P: Billy, could you please start translating the problem?
Billy: We should list our known values.
Mr. P: True, but how specifically should we start
listing our known values?
Bobby: Psst, remember the ball is flying through
the vacuum that you can breathe in, two dimensions.
Billy: The ball is in projectile motion so we should split
our known variables into the x and y directions.
Mr. P: Yes.
(marker squeaking)
Mr. P: Bobby, could you please start listing our known values
in the x and y directions?
Bobby: The displacement in the y direction is 0.70 meters.
Bo: Careful, displacement is a vector,
it has both magnitude and direction.
Bobby: Yeah so, delta y is down, so it should be
negative 0.70 meters.
Mr. P: What else, Bo?
Bo: The acceleration in the y direction is
negative 9.81 meters per second squared.
Mr. P: And, Billy?
Billy: In the y direction the object has a constant acceleration
so we can use the Uniformly Accelerated Motion equation.
It's in free-fall.
Mr. P: Sure, you can use the UAM equations in the y direction,
and in the x direction the ball's moving
at a constant velocity.
Mr. P: I wrote those down, however in the future I'm not going to
because this is how projectile motion works.
Billy: The velocity in the x direction is 10 miles per hour,
but we need to convert that to meters per second, but how?
Bobby: There are 60 seconds in a minute,
and 60 minutes in an hour, and 60 times 60 is 3600.
Therefore, if you multiply by one hour over
3600 seconds, you'll convert to seconds.
There are 1609 meters in a mile, so multiply it
by 1609 meters over one mile, and you'll get
the velocity in the x direction in meters per second.
(marker squeaking)
Bobby: It's 4.4694444 meters per second.
Mr. P: Great, and what else do we know?
Billy: The velocity final of the ball is zero.
Mr. P: Actually no, the velocity final of the ball
does not equal zero.
I do agree that after the ball runs into the bucket
it will stop, however everything we're listing here
has to do with projectile motion while the ball
is flying through the vacuum that you can breathe
in at least two dimensions.
As soon as it touches that bucket,
it's no longer in projectile motion.
Bo: The initial velocity of the y direction
is 10 miles per hour.
Mr. P: Unfortunately, again this is not correct.
The initial velocity of the ball in the y direction
is not 10 miles per hour.
Looking at the video of the ball before it is dropped,
Billy, what can you tell me about the motion of the ball
in the y direction?
Again, before it leaves my hand?
Billy: Oh I see, the ball isn't moving up or down.
Bo: Yeah, so the initial velocity
in the y direction is zero.
Mr. P: Correct, before the ball is dropped,
it isn't moving in the y direction.
Therefore, its initial velocity as it's being dropped
in the y direction is equal to zero.
I also wrote down that the displacement
in the x direction is equal to question mark,
because that's what we're trying to find.
I know it was in our picture, but I figured
it would be good to have it listed
among our givens as well.
We have all our known values listed.
Who can tell me which direction
we should start with, and why?
Bo: We should start in the y direction
because we know three of the variables.
We only know one in the x direction, which means
we can't work with the x direction yet,
because we need to know two variables there.
Mr. P: Yes, we know three variables in the y direction,
and therefore we should start there.
And what, Bobby, are we going to solve for?
Bobby: You just wrote it on the board.
We're trying to find the displacement in the x direction.
Mr. P: Ah yes, you are correct, the displacement
in the x direction is what we're trying to find
in the problem, however we can't actually
solve for that directly.
There's something else we have to solve for first.
Bobby: We're looking for a change in time,
because it's a scalar.
Mr. P: Very nice, yes.
Billy, could you please find the change in time
using the y direction?
Billy: We can use the equation, change in position
in the y direction equals the velocity initial
in the y direction, times the change in time,
plus one half times the acceleration in the y direction,
times the change in time squared.
We can plug in the numbers to get negative 0.7
equals zero times a change in time, plus one half times
negative 9.81 times the change in time squared.
Zero times anything is zero, so that part goes away
and we can multiply the whole equation by two,
which gives us two times negative 0.7 equals
negative 9.81 times a change in time squared.
Divide both sides by negative 9.81 to get,
the change in times squared equals two times
negative 0.7 divided by negative 9.81.
And then we can take the square root
of the whole thing to get the change in time
equals the square root of two times negative 0.7
divided by negative 9.81, which works out to be...
Bo: 0.377772 seconds.
(marker squeaking)
Mr. P: Now that we know how long it will take the ball
to go from my hand to the bucket,
what do we do next?
Bo?
Bo: Well, because change in time is a scalar
and therefore independent of direction,
we can use the change in time in the x direction.
So we should use the equation, velocity in the x direction
equals the change in displacement--
Mr. P: Actually, remember it's change in position
that x means "position" and delta means "change in",
so delta x is the change in position of the object,
or the displacement of the object.
Bo: Right, thank you.
Velocity in the x direction equals the change in position
over the change in time.
Multiply through by the change in time to get
displacement equals velocity in the x direction
times the change in time.
We have numbers, so it's 4.4694444 times 0.377772,
which works out to be 1.68843.
Or with two sig figs, 1.7 meters.
(marker squeaking)
Mr. P: Great, now we know we need to drop the ball
1.7 meters in front of the bucket in order for it
to land in the bucket.
Let's see if it works.
(running footsteps)
[Voiceover] Three, two, one.
(ball smacks bucket, boy shouts)
(blues rock music)
Geneve: You Got It!!
Billy: Actually Mr. P, it looks like you released the ball
a bit after that cone.
You should have released it right at the cone
because the cone is 1.7 meters in front of the bucket.
Bo: Yeah, that is kind of strange.
Bobby: That's probably because of air resistance,
right Mr. P?
Mr. P: Okay, well, yes.
There is air resistance and it does make a difference,
but you might be surprised to learn that
it actually only decreases the displacement
in the x direction of the ball by less than one centimeter.
What actually has a much larger effect on the ball
is the speed of the car, which is a variable
that was actually very difficult to control.
It turns out if you decrease the speed of the car
by half a mile per hour, you decrease the displacement
in the x direction of the ball
by more than eight centimeters.
Which made this a very imprecise experiment,
but it sure was fun.
Trying to drop the ball at the right time
was actually also very difficult. (chuckles)
Billy: Oh okay, that makes sense.
Bo: Okay.
Bobby: Okay.
Mr. P: Of course, we aren't actually done with the problem.
This was just Part One.
What we're going to do in our next lesson
is we are going to determine the final velocity of the ball
right before it strikes the bucket.
Hopefully, I will see you there soon.
(sped-up dialogue)