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# Projectile motion (part 4)

Video transcript

We'll now use that equation we
just derived to go back and solve-- or at least address--
that same problem we were doing before, so let's write
that equation down again. Actually, let's write
the problem down. Lets say I have the cliff
again, and so my initial distance is 0, but it goes
down 500 meters. I'm not going to redraw the
cliff, because it takes a lot of space up on my limited
chalkboard. We know that the change
in distance is equal to minus 500 meters. I'm still going to use the
example where I don't just drop the ball, or the penny, or
whatever I'm throwing off the cliff, but I actually throw
it straight up, so it's going to go up and slow down
from gravity, and then it will go to 0 velocity, and start
accelerating downwards. You could even say decelerating in the other direction. The initial velocity, vi, is
equal to 30 meters per second, and of course, we know that the
acceleration is equal to minus 10 meters per second
squared-- it's because acceleration gravity is always
pulling downwards, or towards the center of our planet. If we wanted to figure out the
final velocity, we could have just used the formula, and we
did this in the last video-- vf squared is equal to
vi squared plus 2ad. Now what I want to do is use the
formula that we learned in the very last video to figure
out-- how long does it take to get to the bottom, and
to hit the ground? Let's use that formula: we
derived that the change in distance is equal to the initial
velocity times time plus acceleration time
squared over 2, and that's initial velocity. The change in distance is minus
500, and that's equal to the initial velocity-- that's
positive, going upwards, 30 meters per second, 30t. I'm not going to write the units
right now, because I'll run out of space, but you can
redo it with the units, and see that the units
do work out. When you square time, you have
to square the time units, although we're solving
for time. Then, plus acceleration, and
acceleration is minus 10, and we're going to divided it by 2,
so it's minus 5t squared. We have minus 500 is equal to
30t plus minus 5t, and we could just say minus
5t squared, and get rid of this plus. At first, you say, Sal-- there's
a t, that's t to the first, and t to the second,
how do I saw solve this? Hopefully, you've taken algebra
two or algebra one, in some places, and you remember
how to solve this. Otherwise, you're about to learn
the quadratic equation, although I recommend you go
back, and learn about factoring in the quadratic
equation, which there are videos on that I've
put on Youtube. I hope you watch those first
if you don't remember. We can do this-- let's put these
two right terms on the left hand side, and then we'll
use the quadratic equation to solve, because I don't think
this is easy to factor. We'll get 5t squared minus 30t
minus 500 is equal to 0-- I just took these terms and put
them on the left side. We could divide both sides by
5, just to simplify things, and so we get t squared minus
6t minus 100 is equal to 0. I could do that, because 0
divided by 5 is just five, so I just cleaned it
up a little bit. Let's use the quadratic
equation, and for those of us who need a refresher,
I'll write it down. The roots of any quadratic--
in this case, it's t we're solving for-- t will equal
negative b plus or minus the square root of b squared minus
4ac over 2a, where a is a coefficient on this term, b is
the coefficient on this term, negative 6, and c is the
constant, so minus 100. Let's just solve. We get t is equal to negative
b-- so negative this term. This term is negative 6, so if
we make it a negative, it becomes plus 6. It becomes 6 plus or minus the
square root of b squared, so it's minus 6 squared, 36,
minus 4 times a, and the coefficient on a is here,
and that's just times 1. With 4ac, c is a constant term,
minus 100, minus 4 times 1 times minus 100, and all of
that is over 2a-- a is 1 agains, so all of
that is over 2. That just equals 6 plus or minus
the square root-- this is minus 4 times minus 100, and
these become pluses, so it becomes 36 plus 400. So, 6 plus or minus
436 divided by 2. This is not a clean number,
and if you type into a calculator, it's something on
the order of about 20.9. We can just say approximately
21-- you might want to get the exact number, if you're actually
doing this on a test, or trying to send something to
Mars, but for our purposes, I think you'll get the point. I'll say it approximately now,
because we're going to be a little off, but just to have
clean numbers, this is approximately 21--
it's like 20.9. We'll say 6 plus or minus--
let me just write 20.9-- 20.9 over 2. Let me ask you a question:
if I do 6 minus 20.9, what do I get? I get a negative number,
and does a negative time make sense? No, it does not, and that means
that somehow in the past-- I don't want to get
philosophical-- the negative time in this context will
not make sense. Really, we can just stick to the
plus, because 6 minus 20 is negative, so there's only one
time that will solve this in a meaningful way. Time is approximately equal to
6 plus 20.9, so that's 26.9 over 2, and that equals
13.45 seconds. That's interesting. I think if you remember way
back, maybe four or five videos ago, when we first did
this problem, we just dropped the penny straight
from the height. Actually, in that problem, I
gave you the time-- I said it took 10 seconds to hit the
ground, and we worked backwards to figure out that the
cliff was 500 meters high. Now, if you're here at the top
of a 500 cliff, or building, and you drop something that has
air resistance-- like a penny, that has very air
resistance-- it would take 10 seconds to reach the ground,
assuming all of our assumptions about gravity. But if you were to throw the
penny straight up, off the edge of the cliff, at 30 meters
per second right here, it's going to take 13.5--
roughly, 13 and 1/2 seconds-- to reach the ground. It takes a little bit longer,
and that should make sense because-- I have time to
draw a little picture. In the first case, I just took
the penny, and its motion just went straight down. In the second case, I took the
penny-- it first went up, and then it went down. It had all the time when it went
up, and then it went down a longer distance, so it makes
sense that this time-- this was 10 seconds, while this time
was 13.45 five seconds. You can kind of say that it
took-- well, you actually can't say that. I don't want to get too
involved, but I hopefully this make sense to you. If you have a smaller number
here, you should have gone and checked your work, because why
would it take less time when I throw the object straight up? Hopefully, that gave you a
little bit more intuition, and you really do have in your
arsenal now all of the equations-- and hopefully, the
intuition you need-- to solve basic projectile problems. I'll now probably do a couple
more videos where I just do a bunch of problems, just to
really drive the points home. Then, I'll expand these
problems to two dimensions and angles. Before we get there, you might
want to refresh your trigonometry. I'll see you soon.