- This is the Nerd-A-Pult.
(intense suspenseful music)
(upbeat music)
Mr. P: Today, we are going to use it to do
an Introductory Projectile Motion Problem.
(ball impacts Bo) ^- Bo: Ow, what was that?
(Billy chuckles fiendishly)
♫ flipping physics ♫
Mr. P: I purchased the Nerd-A-Pult from
Karl and Mary Houser of Ancient Artillery.
And you can get your own at www.marshmallowcatapults.com.
And, as you can see from the title of the website,
it is originally intended to launch marshmallows.
(children laughing and screaming happily)
(playful acoustic music)
Child: First try!
Mr. P: However, because marshmallows are heavily
affected by the force of drag,
I retrofitted mine, using this bottle cap
to hold this ball instead.
Let's get to the problem.
Billy, could you please read the problem?
Billy: Sure, a ball is launched from the Nerd-A-Pult
with an initial speed of 3.25
meters per second, at an angle
of 61.7 degrees above the horizontal.
If the basket is 93 centimeters from the ball horizontally,
where should the basket be placed vertically
relative to the ball so the ball lands in the basket?
So let me get this straight, Mr. P.
We know the displacement in the x-direction
of the ball will be 93 centimeters, and we are trying
to find the displacement of the ball
in the y-direction, right?
Mr. P: Yes Billy, you are correct.
Basically, we know the horizontal displacement
from the ball to the basket is 93 centimeters.
We know the initial velocity of the ball,
and we're trying to figure out where
to place the basket vertically relative to the ball,
such that the ball will land directly in the basket.
And yes, I did get this from my children.
And, in case you're curious, I did make a video of
how I determined the initial velocity off the Nerd-A-Pult.
And you are more than welcome to watch that.
Bo, at this time could you please begin solving the problem?
Bo: Well, the ball is an object in projectile motion.
So we need to list our givens in the x and y-directions.
We know the acceleration of the y-direction
is -9.81 meters per second squared.
We also know the displacement of the y-direction is
what we're trying to find, so delta y equals question mark.
Bobby: And the displacement in the x-direction is 93 centimeters.
Bo: Yeah.
Billy: And we need to convert that to meters,
so multiply it by one meter over 100 centimeters
to get 0.93 meters.
Mr. P: Good, what else?
What about this initial velocity?
Billy: The initial velocity in the x-direction
is 3.25 meters per second.
Bobby: Actually, I think 3.25 meters per second
is the initial velocity in the y-direction.
Bo: No, the initial velocity in the y-direction is 0.
Mr. P: Actually, I'm sorry,
but none of those statements are correct.
Bobby: Components, we need to break
the initial velocity into its components.
Velocity-initial in the x-direction,
velocity-initial in the y-direction.
Bo: Yeah.
Mr. P: Bobby, you are correct.
The initial velocity needs to be broken into its components.
The initial velocity in the y-direction
and the initial velocity in the x-direction.
And, Billy, could you please find
either component of the initial velocity?
Billy: We know sine theta equals opposite over hypotenuse
and opposite theta initial
is the velocity-initial in the y-direction.
And the hypotenuse is the initial velocity.
Multiply the whole equation by the initial velocity
and you get the initial velocity in the y-direction
equals the velocity-initial times
the sine of the initial angle.
We have all those numbers, so it is 3.25 times
the sine of 61.7 degrees.
Which is 2.86155 meters per second.
Mr. P: Now that we know are initial velocity in the y-direction,
I have added that to our y-direction known values.
And, Bobby, could you please now find
the initial velocity in the x-direction?
Bobby: We know cosine theta equals
the adjacent over the hypotenuse,
and adjacent to the initial angle's
the initial velocity in the x-direction
and that hypotenuse is the initial velocity.
Multiply by the initial velocity
and we get the initial velocity in the x-direction
equals the initial velocity times the cosine
of the initial angle, or 3.25
times the cosine of 61.7,
which is 1.54079 meters per second.
Mr. P: Now that we have figured out
the initial velocity in the x-direction,
I have added it to our x-direction known values.
And we can really begin solving the problem.
So, who can tell me which direction we can start with,
how we know we can start with that direction,
and what exactly we are solving for?
Yeah, Bo?
Bo: We know two variables in both directions
and we actually need three to solve in the y-direction,
but we only need two to solve it in the x-direction,
so we should start with the x-direction.
And we're solving for change in time
because delta t is a scalar.
Mr. P: Very good, Bo.
Bobby, could you please begin working on the x-direction?
Bobby: We know the velocity in the x-direction is constant,
so it's not just the velocity-initial in the x-direction,
but rather it's the velocity in the x-direction
for the whole time.
And we know the velocity in the x-direction
equals a change in position over a change in time.
We can multiply both sides by the change in time,
to cancel out change in time on the right-hand side.
And we get the velocity in the x-direction
times the change in time equals the x-displacement.
Now we can divide the whole thing
by the velocity in the x-direction.
Then we get the change in time equals the x-displacement
over the velocity in the x-direction.
Substituting in numbers give us .93
divided by 1.54079, which is--
Billy: 0.603588 seconds.
Mr. P: Now that we have the change in time, which is a scalar,
which means that it's independent of direction,
we can add that to our y-direction knowns.
And solve for the displacement in the y-direction.
Bo?
Bo: Because the acceleration of the y-direction is constant,
we can use the uniformly accelerated motion equations.
So, the y-displacement equals the initial velocity
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.
Plugging in numbers gives us
2.86155 times 0.603588,
plus 1/2 times -9.81, times 0.603588 squared.
Which works out to be...
- 0.059785 meters.
Mr. P: Great, I'm going to multiply our answer
by 100 centimeters over 1 meter
to convert to centimeters and round to two sig fig's.
And we get -6.0 centimeters
for the displacement of the ball
in the y-direction after being launched
by the Nerd-A-Pult and traveling
93 centimeters in the x-direction.
In other words, if the basket
is 93 centimeters away horizontally,
it needs to be placed 6.0 centimeters
vertically below where the ball is initially launched from.
And, it turns out the basket is already
6.0 centimeters below
where the ball is already launched from.
You know what, actually, let's go outside,
where there is more light,
which allows a faster shutter speed
so the ball won't be just a blur.
And we can show it in slow motion.
(ball hits basket) Yeah, woo-hoo!
Oh, it went in but it got rejected!
It bounced back out!
(intense epic music)
Bo: Wow, that actually worked.
Bobby: I like it.
Billy: Good job.
Mr. P: Thank you very much.
Let's do a quick review.
We drew a picture, we listed everything
we knew in the x- and y-directions.
We broke the initial velocity into its components
in the x- and y-directions.
We were then able to use the equation
for constant velocity in the x-direction
to solve for the change in time.
The change in time's a scalar, so we were
also able to use that in the y-direction.
And therefore, we could use a uniformly accelerated
motion equation in the y-direction
to solve for the displacement in the y-direction.
Thank you very much for learning with me today.
I enjoyed learning with you.