IB · Physics
Physics A.1.2
Learn Physics A.1.2 in this free IB Physics video lesson for A.1.2 Equations of motion.
Learn Physics A.1.2 in this free IB Physics video lesson for A.1.2 Equations of motion.
Learn Physics A.1.2 in this free IB Physics video lesson for A.1.2 Equations of motion.
The video focuses on projectile motion as part of the kinematics unit in physics. It outlines key assumptions, such as the presence of constant acceleration due to gravity and the neglect of air resistance, which simplifies the analysis of motion.
Key concepts discussed include:
The video concludes with a sample problem involving the calculation of time for a projectile to reach a target, emphasizing the importance of breaking down the motion into its components for easier problem-solving.
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
00:23
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.