Good morning. We have already solved this problem, and, we need to work
with unit vectors and projectile motion. So, let’s solve it again, only this time,
using unit vectors! ♪ Flipping Physics ♪
Bobby, please read the problem and Bo, please tell me everything we know in terms of unit vectors.
Hold up mr.p, that sounds exactly like the last problem we did.
It’s not the same. It’s very similar.
But not the same. Close enough.
Y’all did that last time. Do you really not learn? - Do you not get a joke?
Uh, I guess not. - Yeah. So, this is actually the
fourth lesson to use the nerd-a-pult. And, just like the second lesson, this fourth lesson solves
the same problem as the lesson before it, only we are using unit vectors. Bobby and Bo, please?
A ball is launched from the Nerd-A-Pult with an initial speed of 3.25 m/s at an angle of 61.7°
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?
Okay. So, I get it now. We know the initial and final position vectors,
the initial velocity vector, and the acceleration vector. Let’s set the initial point at zero.
So, r position naught equals 0 times unit vector i plus 0 times unit vector j meters.
The final position of the ball is 8.7 cm or 0.087 m in the y-direction and an unknown distance,
let’s call it x, in the x-direction. So, final r position equals x times i plus 0.087 times j
meters. The initial velocity is exactly the same as the last time we did this problem.
We need the initial velocity components in the x and y-directions which works out to be again,
exactly the same as last time when we used unit vectors.
The acceleration of any object in projectile motion is 9.81 meters per second squared in
the negative y-direction. And, it has no acceleration in the x-direction so,
zero times i, plus, negative 9.81 times j meters per second squared.
Thank you Bo. Billy, please begin solving the problem.
Absolutely. Like before, let’s use the uniformly accelerated motion equation r position equals r
naught plus velocity naught times time plus one-half acceleration times time squared.
Substitute all of those unit vector equations into that equation. Remove the zeros,
multiply through by time, and multiply through by time squared. Let’s isolate the x-direction terms
and we get all the terms that have unit vector i in them. So, that is x equals 1.54079 times
time. Huh. We have 2 unknowns, so we cannot do anything with this equation right now. Let’s put
that in our equation holster. Equation holster!
Let’s switch over to the y-direction and all the j unit vector terms. We get 0.087 equals 2.86155
times time minus 4.905 times time squared. Oh boy, there is only one unknown, however,
we are going to have to use the quadratic formula. I’m not good with the quadratic formula.
Thanks Billy. Yes, we are going to need to use the quadratic formula. And yes, believe it or not,
students tend to struggle when using the quadratic formula. So, here is my requirement.
Whenever we use the quadratic formula, we have to actually write out the equation, every time.
Every time? Every time.
Every time. Eh.
Bo. Please give it a whirl. Solve for time in this equation.
Sure. Start by subtracting 0.087 from both sides and rearrange the equation from largest
to smallest exponent. The quadratic formula is, because we have to write it out every time.
Negative b plus or minus the square root of … (Just a second Bo.) It is called the quadratic
formula, not the quadratic expression. Therefore, something equals this. What is this equal to?
Okay. Time. Time is the unknown in the quadratic equation; therefore, the quadratic formula solves
for time. So, time equals negative b plus or minus the square root of b squared minus
4 times a times c all divided by the quantity 2 times a. The numbers in front of variables are …
The coefficients. Thanks?
The numbers in front of the variables are called coefficients.
Right. Thanks. The coefficients are ... well, a is negative 4.905, b is 2.86155,
and c is negative 0.087. So, it works out to be negative 2.86155 plus or minus 2.54588 all divided
by negative 9.81. Time then equals 0.291697 plus or minus 0.259519 which equals 0.551216
or 0.0321784 seconds. The quadratic formula gives us two real solutions for time, but,
how do we know which one is correct? Thanks Bo. In a bit, I will explain
how we know it is the larger of the two times. For now, let’s finish the
problem using that larger time. Bobby, please? Okay. uh. Well, we go back to the equation holster or the x-direction or
i unit vector equation and substitute that larger time in. So, x equals 1.54079 times
0.551216 which equals 0.849308 meters or 85
cm, with 2 sig figs. That means the basket needs to be placed 85 cm horizontally from the initial
x-position of the ball, just like what we got last time. Yep, the physics still works.
Cool. Uh huh.
Y’all really do like redundancy. Very nice Bobby. Alright, let’s talk a
little bit more about all the useful information we can glean from using the quadratic formula.
First off, remember that the quadratic equation is an equation which describes a parabola.
This parabola describes the position of the ball as a function of time.
And the quadratic formula gave us this equation for time: In other words, the time to the top of
the parabola is 0.291697 or 0.29 sec. This is the time at which the ball reaches its maximum height.
0.259519 or 0.26 seconds is how long after the ball reaches its maximum height that the
ball reaches a height of 0.087 m. 0.259519 or 0.26 seconds is also how long before the ball
reaches its maximum height that the ball reaches a height of 0.087 m. So you can see, the ball is
at a height of 0.087 m twice after launch, once 0.0321784 or 0.032 seconds after launch while
the ball is going up, and once 0.551216 or 0.55 seconds after launch while the ball is going down.
In order for the ball to “land in the basket” the ball must be moving down, therefore,
it must be the larger of the time times, the second time the ball is at a height of
0.087 meters and the ball is moving downward, that it reaches the basket.
That is how we know the time to use is 0.551216 seconds. Hopefully you now recognize how useful
the quadratic formula can be, and you remember that you must write down the quadratic formula
every time you use it. - Every time.
Every time. Thank you very much for learning with me today, I enjoyed learning with you.
(Yep. Y’all do love your redundancy.)