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# El movimiento de un proyectil (parte 2)

Transcripción del video

In the last video, I dropped
myself or a penny from the top of a cliff. We started off at 0 velocity,
obviously because it was stationary, and at the bottom,
it was 100 meters per second. We used that to figure out how
high the cliff was, and we figured out that the cliff
was 500 meters high. What I want to do now is let's
do that same problem, but let's do in a general form, and
see if we can figure out a general formula for a
problem like that. Let's say that you have the same
thing, and let's say the initial velocity-- you're given
the initial velocity, you're given the final velocity,
you're given the acceleration, and you want to
figure out the distance. This is what you're given,
and you want to figure out the distance. Doing it the exact same way we
did in that last presentation, but now we're now [INAUDIBLE] formulas, we know that the
change in distance is equal to the average velocity times--
we could actually say the change in time, but I'll just
say it with time, because we always assume that we
start with time equals 0-- times time. We know that the average
velocity is the final velocity plus the initial velocity
divided by 2, so that's the average velocity-- let me
highlight-- this is the same thing as this, and then
that times time. What's the time? You could figure out the time
by saying, we know how fast we're accelerating, and we know
the initial and final velocity, so we can figure out
how long we had to accelerate that fast to get that
change in velocity. Another way of saying that, or
probably a simpler way of saying that, is change in
velocity, which is the same thing as the final velocity
minus the initial velocity is equal to acceleration
times time. If you want to solve for time,
you could say the time-- if I just divide both sides of this
equation by a-- is equal to vf minus vi divided by a. We could take that and
substitute that into this equation, and remember-- this
is all change in distance. We say change in distance is
equal to-- let me write this term in yellow-- vf
plus vi over 2. Let me write this
term in green. That's times vf minus
vi over a. Then if we do a little
multiplying of expressions on the top-- you might have
recognized this-- this would be vf squared minus vi squared,
and then we multiply the denominators over 2a. So the change in distance is
equal to vf squared minus vi squared over 2a. That's exciting-- let me
write that over again. The change in distance is equal
to vf squared minus vi squared divided by 2
times acceleration. We could play around with this a
little bit, and if we assume that we started distance is
equal to 0, we could write d here, and that might
simplify things. If we multiply both sides by
2a, we get-- and I'm just going to switch this to
distance, if we assume that we always start at distances
equal to 0. di, or initial distance,
is always at point 0. We could right 2ad-- I'm just
multiplying both sides by 2a-- is equal to vf squared minus vi
squared, or you could write it as vf squared is equal
to vi squared plus 2ad. I don't know what your physics
teacher might show you or written in your physics book,
but of these variations will show up in your physics book. The reason why I wanted to show
you that previous problem first is that I wanted to show
you that you could actually figure out these problems
without having to always memorize formulas and resort
to the formula. With that said, it's probably
not bad idea to memorize some form of this formula, although
you should understand how it was derived, and when
to apply it. Now that you have memorized it,
or I showed you that maybe you don't have to memorize
it, let's use this. Let's say I have the
same cliff, and it has now turned purple. It was 500 meters high-- it's
a 500 meter high cliff. This time, with the penny,
instead of just dropping it straight down, I'm going to
throw it straight up at positive 30 meters per second. The positive matters, because
remember, we said negative is down, positive is up-- that's
just the convention we use. Let's use this formula, or any
version of this formula, to figure out what our final
velocity was when we hit the bottom of the ground. This is probably the easiest
formula to use, because it actually solves for
final velocity. We can say the final velocity
vf squared is equal to the initial velocity squared-- so
what's our initial velocity? It's plus 30 meters per second,
so it's 30 meters per second squared plus 2ad. So, 2a is the acceleration of
gravity, which is minus 10, because it's going down, so it's
2a times minus 10-- I'm going to give up the units for
a second, just so I don't run out of space-- 2 times minus
10, and what's the height? What's the change in distance? Actually, I should be correct
about using change in distance, because it matters
for this problem. In this case, the final distance
is equal to minus 500, and the initial distance
is equal to 0. The change in distance
is minus 500. So what does this get us? We get vf squared is equal to
900, and the negatives cancel out-- 10 times 500 is 5,000, and
5,000 times 2 is 10,000. So vf squared is equal
to 10,900. So the final velocity is equal
to the square root of 10,900. What is that? Let me bring over my trusty
Windows-provided default calculator. It's 10,900, and the
square root. It's about 104 meters per
second, so my final velocity is approximately-- that
squiggly equals is approximately-- 104
meters per second. That's interesting. If I just dropped something--
if I just drop it straight from the top-- we figured out
in the last problem that at the end, I'm at 100
meters per second. But this time, if I throw it
straight up at 30 meters per second, when the penny hits
the ground, it's actually going even faster. You might want to think about
why that is, and you might realize it. When I throw it up, the highest
point of the penny-- if I throw it up at 30 meters
per second, the highest point of the penny is going to be
higher than 500 meters-- is going to make some positive
distance first, and then it's going to come down, so it's
going to have even more time to accelerate. I think that makes some
intuitive sense to you. That's all the time I have
now, and in the next presentation, maybe I'll use
this formula to solve a couple of other types of problems.