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.
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 93 cm from the ball horizontally, where should
the basket be placed vertically relative to the ball so the ball lands in the basket?
What am I doing? - You are giving me everything we
know in terms of unit vectors. - I do not know how to do that.
Okay. I’ll get us started with the initial and final positions.
Because these positions are in multiple dimensions at once, we use the symbol r instead of x, y,
or z. So, the initial position, or r naught, is well, we get to choose that, so let’s choose
zero i plus zero j as the initial position of the ball. The ball starts at zero. The final position,
which is r final, equals 93 cm in the x direction, which we know is 0.93 m in the x direction,
or 0.93 times unit vector i. The final position in the y-direction is the unknown we are solving for,
so let’s substitute in an unknown variable y, or y times unit vector j. So, the final
position equals 0.93 i plus y j meters. Alright Bo, your turn. Please determine the initial
velocity of the ball in terms of unit vectors. Oh kay. Uh. Well, the initial velocity given
in the problem is in terms of magnitude and direction, so,
we are going to need the x and y-direction components of it. I think I know how to do this.
The initial velocity, or v naught, equals the x component of the initial velocity times i
plus the y component of the initial velocity times j. The x component of the initial velocity equals,
well, because theta is with the horizontal, it equals initial velocity times cosine theta.
The y component of the initial velocity equals initial velocity times sine theta. Initial
velocity is 3.25 meter per second and theta is 61.7 degrees. So, the initial velocity is
1.54079 i plus 2.86155 j meters per second. Very nice Bo. Billy, please give me the
acceleration of the ball using unit vectors. Absolutely. Okay, the acceleration of any object
in projectile motion is 9.81 meters per second squared down. So, objects in projectile motion
have no acceleration in the x-direction so, zero times unit vector i, plus, well negative 9.81
times unit vector j meters per second squared. Thank you Billy. Now, we know the acceleration
of the ball is constant, so we can use the uniformly accelerated motion equations.
Let’s use position final equals initial position plus initial velocity times time plus one-half
acceleration times time squared. Notice we can substitute the unit vector versions
we just determined of final position, initial position, initial velocity, and acceleration.
And we can remove the zeros and multiply through by time and time squared. Bobby,
please isolate all of the x-direction terms and see what we can accomplish with that.
Sure. Um. Isolating all of the terms which have a unit vector i in them gives us 0.93 equals 1.54079
times time. Oh, we can solve for time. Time equals 0.93 over 1.54079 or 0.603588 seconds.
Thanks Bobby. Can you actually also do the same thing with the y-direction?
Yeah. Isolating all the terms which have unit vector j in them gives us y equals 2.86155
time minus 4.905 time squared. And we just figured out the time. So, substitute that in and we get
negative 0.0599785 meters or well, that is negative 6.0 cm. In other words,
the basket needs to be placed 6.0 cm below the original height of the ball.
Just like we got last time. Yep, the physics still works.
Cool. Uh huh.
Very nice Bobby. Now, notice how, by using unit vectors, we are able to combine everything we
know in this problem into one equation. In the long run, that can be very useful.
Thank you very much for learning with me today, I enjoyed learning with you.