Mr. P: Good morning.
Today we're going to continue working on
an introductory projectile motion problem
where the object has an initial velocity
that is entirely horizontal.
This time we are going to determine the final velocity
of the ball right before it lands in the bucket.
(dramatic music) Here we go.
Bo: Hey guys.
Billy: Hey Bo.
Bobby: Hi, Bo.
Mr. P: Last time we determined
how far in front of the bucket to drop the ball
such that it would land into the bucket.
I'm not going to review that now,
so if you didn't watch that video I suggest
that you do so, now.
Billy, could you please help us find
the final velocity of the ball right before
it lands in the bucket?
Billy: Well, we can use the UAM equation,
velocity final squared equals velocity--
velocity final squared equals velocity-- - Mr.P: Billy, are you talking about
the x direction or the y direction?
Billy: Sorry, velocity final in the y direction squared,
equals velocity initial in the y direction squared,
plus two times the acceleration in the y direction
times the change in the y position.
We know that the initial velocity in the y direction is zero
and the acceleration in the y is negative 9.81
and the displacement in the y direction is negative 0.7.
Taking the square root, we get to the final velocity
in the y direction equals the square root
of two times negative 9.81 times negative 0.7,
or...
3.70594.
With two sig figs, that works out to be
3.7 meters per second.
Bobby: Actually, remember you need to be
smarter than your calculator.
Any time you're taking the square root,
you need to set up whether it's positive or negative
because its position could be positive or negative.
Because the ball is going down,
it's a negative 3.7 meters per second.
Billy: Stupid calculator.
(marker squeaking)
Mr. P: So the velocity of the ball right before
it lands in the bucket is 3.7 meters per second, down.
Bo: Mr. P?
Mr. P: Yes, Bo?
Bo: Isn't that just the velocity final in the y direction?
Don't we need to include
the velocity in the x direction too?
Mr. P: Absolutely.
(marker squeaking)
Notice that the final velocity we're solving for
is the hypotenuse of the triangle, or the resultant vector.
Where the final velocity in the y direction
and the final velocity in the x direction
are the components of the resultant vector.
They are the components of the final velocity.
Will you look at that?
We get to review component vectors.
How awesome is that?
Billy: That is awesome.
Bo: No.
Mr. P: Okay, let's see, we know the final velocity
in the y direction, so we just need
one more side of the triangle,
we need the final velocity in the x direction.
Bo, please help us out.
Bo: Actually, we already know it.
The ball is travelling at a constant velocity
in the x direction, so it's always
4.4694444 meters per second.
Mr.P: Correct.
Because the velocity in the x direction is constant,
it is the initial velocity, the final, the middle,
it doesn't matter, it's always the same because
that velocity in the x direction is constant.
Bobby, now that we have two sides of the triangle,
please figure out the final velocity.
Bobby: Well, we have two sides of a right triangle
so we can use a squared plus b squared equals c squared,
or velocity final squared equals the velocity final
in the x direction squared plus the velocity final
in the y direction squared.
Taking the square root of both sides gives us,
that the velocity final equals the square root
of velocity final x squared plus
the velocity final y squared.
With numbers, that is the square root of 4.4694444 squared
plus the quantity negative 3.70594 squared.
And that works out to be...
Bo: Positive or negative 5.80602, with sig figs,
negative 5.8 meters per second.
(marker squeaking)
Billy: Actually it should be positive
5.8 meters per second because the ball
is moving to the right.
Bo: Actually the ball is moving down,
so the final velocity is negative.
Billy: No, it's positive.
Bo: Negative.
Billy: No, the final velocity is positive.
Bo: Negative!
Billy: Cheeky gormless nimwit!
Bo: Ill-fated buffoon!
Bobby: You know, you're both wrong.
Velocity is a vector.
We only figured out the magnitude.
We need to figure out the direction.
We need to figure out theta!
You're both wrong.
Mr. P: Wow, okay.
Bobby, thank you.
You are correct, and I think I'll figure out the angle
just to give you two some time to calm down.
(marker squeaking)
Billy: Ill-fated buffoon?
Bo: Cheeky gormless nimwit?
Bobby: I think both are equally applicable.
Mr. P: In a right triangle, the sign of the angle theta
equals the opposite over the hypotenuse.
Opposite our angle theta is the velocity final in the
x direction, on the hypotenuse we have the velocity final.
Take the inverse sine of both sides, we get that
theta is equal to the inverse sine of velocity final
in the x direction divided by the velocity final.
So that works out to be, with numbers,
the inverse sine of 4.4694 repeating, divided by 5.80602.
And what do we get?
Billy: 50.3354 degrees.
(marker squeaking)
Mr. P: Therefore, with two significant digits,
the final velocity is 5.8 meters per second
at an angle of 50 degrees in front of
the negative y axis.
50 with two sig figs, in order to illustrate that,
it needs to be 5.0 times 10 to the first,
which is a bummer but that's probably
the best way to illustrate it.
Notice, in front of the negative y axis,
we can't use cardinal directions here like east or south
because this isn't north, south, east, west,
this is up, down, left, right.
In order to understand that, let's draw an axis
on the tail of our resultant vector.
You can see that the angle theta
is relative to the negative y axis.
We say in front of the negative y axis
to indicate that that's the direction the ball is moving.
Notice that the initial velocity of the ball
is about 4.47 meters per second, horizontally to the right.
As the ball falls, the magnitude of its velocity increases
and the direction of the velocity points
farther and farther down.
Until right before it runs into the bucket,
and has a velocity of approximately 5.81 meters per second
at an angle of approximately 50.3 degrees
in front of the negative y axis.
Billy: Oh, I get it.
The velocity vector arrows increase in length
as the ball falls, because the magnitude
of the ball's velocity keeps increasing.
Bo: Yep.
Mr. P: And I have one more question.
As you can see in the video, my hand is actually
always directly above the ball.
Why is that?
Why is my hand always directly above the ball?
Bo: That's because the car keeps moving
at a constant 4.694444 meters per second,
and therefore the car and the ball
will have the same displacement in the x direction.
Mr.P: Exactly correct.
The car, and therefore my hand, and the ball,
both move at about a constant horizontal velocity
of 4.47 meters per second.
Therefore, both will always have the same x position.
And notice that the ball appears to move straight down
from the perspective of my hand.
Billy: But wait a moment.
Mr. P, doesn't the ball move farther than your hand?
Mr. P: Yes Billy, you are correct.
The ball does move farther.
The ball moves in both the x and y directions,
and therefore moves in a parabola.
My hand only moves in the x direction,
and therefore only moves in a straight line.
So while their displacements in the x direction
are the same, because their displacements in the y direction
are different, their distance travelled will be different.
Therefore, yes, the ball ends up travelling farther.
Okay.
Please remember, with projectile motion,
that you have to list what you know in the x direction,
list what you know in the y direction,
because the motions are different.
In the x direction you have constant velocity,
in the y direction you have uniformly accelerated motion.
Thank you very much for learning with me today.
I enjoyed learning with you.