Mr. P: Good morning.
Today we're going to use the Nerd-A-Pult
to do another projectile motion problem.
Bo: Hey guys.
Billy: Hey Bo!
Bobby: Hi Bo.
♫ Flipping physics ♫
Mr. P: Bobby, could you
please read the problem?
Bobby: A ball is launched from the Nerd-A-Pult
with an initial speed
of 3.25 m/s at an angle
of 61.7 degrees above the horizontal.
If the basket is 8.7 cm
above the ball vertically,
where should the basket be placed
horizontally relative to the ball
so the ball lands in the basket?
Bo: Hold up, Mr. P.
That sounds exactly like the last
projectile motion problem we did.
Billy: It's not the same.
Bo: It's very similar.
Billy: But not the same.
Bo: Close enough.
Mr. P: Actually Bo, that's why I
worded the problem this way,
to make it look almost exactly the same.
And it turns out that the
physics approach is the same,
however the math ends
up being a little bit
more difficult with this problem.
In the previous problem,
we knew the displacement
in the x direction and we were solving
for the displacement in the y direction.
In this problem, it is reversed.
We know the displacement in
the y direction is 8.7 cm,
and we are solving for the displacement
in the x direction.
Let's draw our picture.
Billy, what should we do now?
Billy: We need to resolve the
initial velocity vector
into its components.
Mr. P: True, using the fact that we know
the magnitude and direction
of the initial velocity,
we need to figure out the initial velocity
in the y direction and
the initial velocity
in the x direction.
And because this is
actually exactly the same
as what we did in our last
projectile motion problem,
I'll just write it out on the board.
And Bo, what should we do now?
Bo: Well, the ball is in projectile motion
so we need to list what we know
in the x and y directions.
Bobby: And we just figured out
the initial velocities.
And the initial velocity in
the x direction is the same
as the constant velocity
in the x direction,
because in projectile motion,
velocity doesn't change
in the x direction.
Billy: And we know the displacement
in the y direction
is 8.7 cm which we
multiply by 1 m over 100cm
to convert to meters, which is 0.087 m.
Bo: And we know the acceleration
in the y direction
is -9.81 m/s squared.
And we're looking for the
displacement in the x direction,
so delta x equals question mark.
Mr. P: Yes, those are all of our known values.
Bobby, what do we do now?
Bobby: Well, we know three
variables in the y direction.
So let's start there to
find our change in time
and then use that in the x direction.
Mr. P: Bobby, that
sounds like a great idea.
Go for it.
Bobby: So, we can use the uniformly
accelerated motion equation.
Displacement 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.
With numbers, that is 0.087
equals 2.86155 times the change in time,
plus one half times -9.81
times the change in time squared.
Aw, man!
How do we solve for change
in time in that equation?
Billy: Oh, we need to use
the quadratic formula.
And I have a program in my calculator
we can use to solve that.
Mr. P: Hold up, Billy.
You are correct that we do need to use
the quadratic formula
to solve this equation.
Now, we don't need the quadratic formula
to solve this problem, and I'll show you
how not to use the quadratic
formula in just a minute.
But sure, we can start
with the quadratic formula.
But no, we're never going to just
"plug it into a program
in our calculator".
We need to make sure that we show
that we understand every single step.
So, let's show every single step.
And let's start by
rearranging this equation
to make it equal to zero.
Bo, please continue.
Bo: We can subtract 0.087 from both sides
to get -4.905 times the
change in time squared,
plus 2.86155 times the change in time
minus 0.087 equals zero.
The quadratic formula would then be
the change in time equals -b plus or minus
the square root of the quantity b squared
minus 4ac all divided by 2a.
Plugging in numbers gives us--
Mr. P: Uh, Bo?
What are the values of a, b, and c?
Bo: Okay, the formulaic quadratic equation
for the quadratic formula would be
a times change in time squared
plus b times change in
time plus c equals zero.
So a is -4.905,
b is +2.86155,
and c is -0.087.
And we can plug all those numbers
into the quadratic formula and we get...
0.551216,
or 0.0321784 seconds.
Billy: How does he do that?
Mr. P: But how do we know which one is correct?
Bo, the change in
time for the ball to go
from here to there clearly is not as small
as one thirtieth of a second.
Therefore the change in
time is 0.551216 seconds.
Looking at all this math on the board,
realize that this is not the way
that I suggest that
you solve this problem.
I'm actually now going to erase all this
quadratic formula math off of the board.
If you want to take a closer look at it,
you're more than welcome to.
My lecture notes are
always posted on my website
so you can look there.
So we're not going to use
the quadratic formula,
we're going to solve
it in a different way.
So the issue is that
everybody wants to solve
for the change in time directly.
But what you should actually do is solve
for the final velocity
in the y direction first.
Billy, could you please do so?
Billy: Well, we know the final velocity
in the y direction squared equals
the initial velocity in
the y direction squared
plus two times the
acceleration in the y direction
times the displacement in the y direction.
Taking the square root
of the whole equation
gives us the final
velocity in the y direction
equals the square root
of the initial velocity
in the y direction squared,
plus two times the
acceleration in the y direction
times the displacement in the y direction.
With numbers, that is the square root
of 2.86155 squared
plus two times -9.81 times 0.087,
which equals...
Bobby: 2.54588 m/s.
Bo: Remember, be smarter
than your calculator.
Bobby: Yeah, that's right.
The solution to a square root can be
either positive or negative,
so positive or negative 2.54588 m/s.
And it is--
Billy: Negative, because we
know the ball is going
down when it lands in the basket,
so -2.54588 m/s.
Mr. P: Yes, and now we need the change in time.
Bo?
Bo: Final velocity y
equals initial velocity y
plus acceleration y
times the change in time.
We need to solve for the change in time,
so subtract the initial
velocity y from both sides
to get final velocity y
minus initial velocity y
equals acceleration y
times the change in time.
Divide both sides by the acceleration y,
and we get the change in time equals
the final velocity y minus
the initial velocity y
divided by the acceleration y,
or -2.54588
minus 2.86155
divided by -9.81, which is--
Bobby: 0.55122 seconds,
which is exactly what we got before.
Mr. P: Correct, and notice that there
is only one change in time.
So unlike when we use
the quadratic formula,
we don't have two and try to figure out
which one is correct.
Actually there are two steps,
but even though there are two steps
there's actually less math than
when we use the quadratic formula.
And we actually know more information.
We know the final velocity
in the y direction as well.
So the only thing left
to do is to figure out
the displacement in the x direction
such that the ball will
land in the basket.
Bobby, could you please figure that out?
Bobby: The velocity in the x direction
equals the displacement in the x direction
divided by the change in time.
And because change in time is a scalar,
it is the same in both
the x and y directions.
So we can use the change in time
we just found in the y direction.
Multiply by the change in time to get
the displacement in the x direction
equals the velocity in the x direction
times the change in time,
or 1.54079
times 0.55122,
or 0.849314 m.
Mr. P: Let's convert to
centimetres by multiplying
by 100 cm over 1 m and
round to two sig figs.
And we get 85 cm for the
displacement in the x direction.
Let's see if it works.
And again, let's go outside
where there is more light
so we can have a faster shutter speed,
1/3000th of a second, if you are curious.
And therefore the ball will not be a blur
and we can show it in slow motion.
Billy: I like that!
Bo: Impressive.
Bobby: Now that is precision.
Or, accuracy.
Bo: It's both.
Mr. P: Thank you.
Now, I do want to point out that
even though this problem looks very
similar to the one we did before,
mathematically it was a
little bit more complicated
because we had to start
out in the y direction
where the change in time is squared,
and that just complicates
things a little bit.
So, let's review.
We started out by breaking
our initial velocity
into its components, then we listed
everything we knew in
the x and y directions.
We started with the y direction
and we used uniformly
accelerated motion equations
to solve for first the final
velocity in the y direction,
then the change in time.
Then we switched to the x direction
and used the change in
time in the x direction
in the equation for constant velocity
to solve for the displacement
in the x direction.
That is what we did.
Thank you very much for
learning with me today,
I enjoyed learning with you.
