00:00Hello, I'm Diego from revision
00:01deja and welcome to the
00:03second video on 8 .1
00:05kinematics. This video will focus
00:07on projectile motion. We have
00:09already seen the entirety of
00:118 .1 .1 and right
00:13now we're going to focus
00:14on 8 .1 .2. Alright,
00:16there is only one learning
00:18objective for this session. So
00:20just write this down and
00:21make sure that at the
00:21end of the lesson you
00:23understand what it says. First,
00:24let's start with a very
00:25quick assumptions section.
00:28Here in projectile motion, in
00:29general NIB physics, we assume
00:31that there is a constant
00:32acceleration acting on all bodies
00:34close to the surface of
00:35the Earth. The Earth is
00:37actually not perfectly around. It's
00:40a little bit flatter at
00:41the poles, so it looks
00:43a little bit more like
00:44this, if we really exaggerate
00:46it. So gravity is actually
00:47even everywhere on Earth, but
00:49we ignore that in IB
00:51physics. Furthermore, we assume that
00:54the surface of the Earth
00:55is large enough
00:56that at any point it
00:58is considered flat. So when
01:00we're looking at a projectile
01:02motion problem, we're assuming that
01:04the surface of the earth,
01:05so the ground is always
01:07flat. All right, let's move
01:08on. So to the projectile
01:10motion, which is the only
01:12projectile motion we care about
01:13in IB physics, there are
01:15a couple typical situations in
01:17projectile motion. First, an item
01:20flying straight up, which is
01:22something we saw in one
01:23of the sample
01:24questions in the previous unit
01:25and also an object going
01:28at an angle. This is
01:31what we call ballistic motion
01:32and essentially the object is
01:34going to have a trajectory
01:36approximately like this. It's very
01:39important that we also consider
01:40that most of the time
01:41we will ignore air resistance
01:43when we're talking about projectile
01:45motion. Let's take a look
01:47at a couple sketches of
01:49what projectile motion graph
01:52looks like. In projectile motion
01:54at displacement time graph would
01:56look something like this. If
01:59the object is thrown upwards,
02:01the object goes up away
02:03from its starting position. This
02:05is the point where it
02:06is the farthest and then
02:07it comes back down. Alright,
02:09so this is for displacement
02:10time graphs for when you
02:13throw an object straight up.
02:15However, the distance time graph
02:17is going to look completely
02:18different because
02:20It's going to look a
02:21little bit more like this.
02:24Indeed, this distance is the
02:26sum of the length of
02:27the path taken. That means
02:29that here is the maximum
02:32height point. So the object
02:33slows down as it reaches
02:35its maximum height and then
02:37accelerates once again as it
02:39comes down. Now we need
02:41to calculate the motion in
02:43both vertical and horizontal planes.
02:45Gravity only acts on the
02:47vertical plane.
02:48So most of the time
02:49you're mostly going to care
02:51about the vertical plane because
02:52we are ignoring a resistance.
02:54So horizontal motion, most of
02:57the time, will have an
02:59acceleration of zero. An important
03:01thing about horizontal motion is
03:03you can calculate the horizontal
03:05distance traveled. So if you're
03:07firing an object out of,
03:09for example, let's say cannon
03:10here, and you want to
03:12know how far it traveled,
03:14you have to do horizontal
03:16total velocity times the time
03:19it took to land. Now
03:21you might ask how do
03:22we find out the horizontal
03:24velocity? Well this is where
03:26vectors and scalars come back.
03:29So let's assume we have
03:30a ball that we throw
03:31at a certain angle which
03:33we're going to say is
03:34a 45 degree angle. We
03:38throw this ball with a
03:39velocity of 10 meters per
03:43second.
03:44And here we want to
03:45know what is the vertical
03:46and what's the horizontal motion.
03:48Well, what we do is
03:50we use trigonometry. Because any
03:53vector quantity and velocity of
03:55vector quantity can be split
03:56into its vertical and horizontal
03:58components. Let's assume this red
04:01arrow is our velocity vector.
04:05We can find its vertical
04:08component and its horizontal component
04:12using
04:12trigonometry. Luckily, we have this
04:1445 degree angle to help
04:15us out to find the
04:17components. We're going to call
04:19this horizontal component the h
04:21and vertical component the v
04:24using trigonometry. We can find
04:26the vh by using cos
04:29of 45 degrees equals vh
04:31over 10 because vh is
04:33the adjacent there and 10
04:35is the hypothenyl. Here, because
04:37we're using 45 degrees, the
04:40vertical and horizontal
04:40components are going to be
04:41equal because 45 is exactly
04:44half of 90. So the
04:46angle here is also going
04:49to be 45 degrees. However,
04:51in questions, it doesn't tend
04:53to be the case and
04:54you have to use cos
04:54of a different angle for
04:56the other component. Let's do
04:57the math. So 10 times
04:59cos 45 equals five square
05:02root two, which is approximately
05:057 .07 meters per second.
05:08That's how you break down
05:10velocity into its components. Let's
05:12see some other quick facts
05:13about projectile motion. If your
05:15motion is in a bell
05:16trajectory, your highest point is
05:18going to be at the
05:18middle. And it's going to
05:20act as an acts of
05:21symmetry for the motion. So
05:23there's going to be the
05:24same horizontal distance between the
05:27point of max height and
05:29the beginning of the motion
05:30and the point of max
05:31height and the end of
05:33the motion. These are going
05:34to be equal. It's the
05:35same for time. The time
05:36taken to the point of
05:38max height is half of
05:39the time for the total
05:41motion. Keep those facts in
05:42mind they will really help
05:44you to solve projectile motion
05:46questions. Alright, that is nearly
05:48it. We just need to
05:50quickly talk about vertical motion
05:51a little bit more. Your
05:53vertical velocity is the most
05:55important factor in your projectile
05:57motion because your vertical velocity
06:00is the only one that
06:00has an acceleration. So your
06:02VV is going to decrease
06:04until the point
06:04of max height. At the
06:06point of max height, and
06:08this is important, so at
06:10t, vv equals zero, because
06:14it's just about to start
06:16moving downwards, and it has
06:18finished its upwards motion. This
06:20is important because it can
06:21help you find that point
06:23of max height. Then also
06:24remember that there's always a
06:27minus 9 .1 acceleration downwards.
06:30It's negative because it's acting
06:32against
06:33vertical velocity. So you must
06:34keep that in mind. Also,
06:36your VV is going to
06:37be negative. It's going to
06:39be smaller than zero in
06:41the second half of the
06:42motion. So, F30. Big T
06:45is the time to maximum
06:46height. All right, I don't
06:47want to complicate projectile motion
06:49more for you because it
06:51is actually a relatively simple
06:52concept if you manage to
06:54understand all of those little
06:55facts. All right, let's look
06:58at our sample question to
06:59finish off this lesson. As
07:00students try to
07:01a tennis ball at point
07:02P. The ball is initially
07:03directed at an angle of
07:047 degrees to the horizontal.
07:06Calculate the time it takes
07:07for the ball to reach
07:09the net. Alright, let's highlight
07:10some important information, so we're
07:12looking for time, we have
07:13an angle of 7 degrees,
07:15and we start at point
07:16P. We have a beautiful
07:18diagram here. And like any
07:21projectile motion question, let's start
07:23by finding the components of
07:26this velocity vector right here.
07:28So our horizontal
07:29and vertical velocities. To do
07:31so, we simply use a
07:32formula as we used before.
07:34So here we need cos
07:367 for our horizontal velocity
07:39and then we need cos
07:42of 90 minus 7 for
07:44our vertical velocity. So let's
07:47find out. Alright, here make
07:50sure to always keep your
07:52answers on round. If you
07:53round them, your final answer
07:55might be wrong. So we
07:56have our vertical
07:57and horizontal velocity is now
07:59what we're looking for is
07:59the time it takes for
08:00the ball to reach the
08:02net. The easiest way to
08:03do this is simply using
08:05horizontal velocity because we know
08:07how far we are from
08:09the net. We know our
08:10horizontal velocity and we know
08:12that our horizontal velocity is
08:13not affected by any acceleration.
08:15So it's actually as far
08:17as forward as doing this
08:19divided by our horizontal velocity.
08:25Don't forget the seconds at
08:26the end and that's it.
08:28It's very quick. Very straightforward,
08:30easy question. Always start with
08:33splitting the components. Then you'll
08:34be able to answer any
08:35question much more easily. All
08:38right, that's it for projectile
08:39motion. I hope you understood
08:40everything otherwise. Please drop us
08:42a comment for the rest
08:43of 8 .1, which is
08:45movement through fluids.