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Projectile motion graphs

Visualizing position, velocity and acceleration in two-dimensions for projectile motion.

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  • leafers seed style avatar for user Xinlu Wen
    Why is the second and third Vx are higher than the first one?
    (14 votes)
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    • purple pi purple style avatar for user Normunds Palabinskis
      All thanks to the angle and trigonometry magic.
      After looking at the angle between actual velocity vector and the horizontal component of this velocity vector, we can state that:
      1) in the second (blue) scenario this angle is zero;
      2) in the third (yellow) scenario this angle is smaller than in the first scenario.

      If above described makes sense, now we turn to finding velocity component. We do this by using cosine function: cosine = horizontal component / velocity vector. After manipulating it, we get something that explains everything!
      horizontal component = cosine * velocity vector

      Now we get back to our observations about the magnitudes of the angles. If we work with angles which are less than 90 degrees, then we can infer from unit circle that the smaller the angle, the higher the value of its cosine.
      So from our derived equation (horizontal component = cosine * velocity vector) we get that the higher the value of cosine, the higher the value of horizontal component (important note: this works provided that velocity vector has the same magnitude. In this case, this assumption (identical magnitude of velocity vector) is correct and is the one that Sal makes, too).

      Now let's get back to our observations:
      1) in blue scenario, the angle is zero; hence, cosine=1. This means that the horizontal component is equal to actual velocity vector. It actually can be seen - velocity vector is completely horizontal.
      2) in yellow scenario, the angle is smaller than the angle in the first (red) scenario. This means that cos(angle, red scenario) < cos(angle, yellow scenario)! Hence, the horizontal component in the third (yellow) scenario is higher in value than the horizontal component in the first (red) scenario.
      (3 votes)
  • orange juice squid orange style avatar for user Prasoon Pandey
    At in the top diagram, shouldn't the ball have a little positive acceleration as if was in state of rest and then we provided it with some velocity?
    (7 votes)
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    • blobby green style avatar for user Beibei
      maybe have a positive acceleration just before into air, once the ball out of your hand, there will be no force continue exerting on it, except gravitational force (assume air resistance is negligible), so in the whole journey only gravity affect acceleration.
      (9 votes)
  • blobby green style avatar for user Mike Phillips
    Why did Sal say that v(x) for the 3rd scenario (throwing downward -orange) is more similar to the 2nd scenario (throwing horizontally - blue) than the 1st (throwing upward - "salmon")? I would have thought the 1st and 3rd scenarios would have more in common as they both have v(y)>0.
    (6 votes)
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  • blobby green style avatar for user Keerthivasan
    how the velocity along x direction be similar in both 2nd and 3rd condition?
    (that is in blue and yellow)
    (4 votes)
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  • marcimus pink style avatar for user pouravibanerjee
    why is the acceleration of the y-value negative? Why is the acceleration of the x-value 0.
    (3 votes)
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  • duskpin ultimate style avatar for user Adara
    At , how is the blue graph's x initial velocity a little bit more than the red graph's x initial velocity?
    (5 votes)
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    • leaf green style avatar for user Ishika Walia
      In this case/graph, we are talking about velocity along x- axis(Horizontal direction)
      So, initial velocity= u cosӨ
      For blue ball and for red ball Ө(angle with which the ball is projected) is different(it is 0 degrees for blue, and some angle more than 0 for red)

      Now, assuming that the two balls are projected with same |initial velocity| (say u), then the initial velocity will only depend on cosӨ in initial velocity = u cosӨ, because u is same for both.

      Now, let's see whose initial velocity will be more -
      It'll be the one for which cos Ө will be more
      For blue, cosӨ= cos0 = 1
      For red, cosӨ= cos (some angle>0)= some value,say x<1
      [because we know that as Ө increases, cosӨ decreases
      cos(0)=1
      therefore, cos(Ө>0)=x<1]

      Now, we have,
      Initial velocity of blue ball = u cosӨ = u*(1)= u
      Initial velocity of red ball = u cosӨ = u*(x<1)= some value, say y<u
      Therefore, initial velocity of blue ball> initial velocity of red ball
      Hence, Sal plots blue graph's x initial velocity(initial velocity along x-axis or horizontal axis) a little bit more than the red graph's x initial velocity(initial velocity along x-axis or horizontal axis).
      Hope this made you understand!
      (1 vote)
  • piceratops tree style avatar for user Zeeshan Hassan
    in the first graph of the second row (Vy graph) what would I have to do with the ball for the line to go upwards into the 1st quadrant? the person who through the ball at an angle still had a negative velocity. second question. if the ball hit the ground an bounced back up, would the velocity become positive?
    (4 votes)
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  • blobby green style avatar for user J
    Hi there, at why does Sal draw the graph of the orange line at the same place as the blue line? I thought the orange line should be drawn at the same level as the red line. Thanks!
    (2 votes)
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  • aqualine ultimate style avatar for user sarikap09609
    Throughout the video, how exactly does the graph that Sal is using in the video?
    (2 votes)
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  • mr pink red style avatar for user Nikhil Sharma
    At the x~t graph is trying to say that the projectile at an angle has the least horizontal displacement which is wrong. The x~t graph should have the opposite angles of line, i.e. the pink projectile travels furthest then the blue one and then the orange one.
    (2 votes)
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    • blobby green style avatar for user Ben
      If the graph was longer it could display that the x-t graph goes on (the projectile stays airborne longer), that's the reason that the salmon projectile would get further, not because it has greater X velocity.
      (1 vote)

Video transcript

- [Instructor] So in each of these pictures we have a different scenario. We have someone standing at the edge of a cliff on Earth, and in this first scenario, they are launching a projectile up into the air. In this one they're just throwing it straight out. They're not throwing it up or down but just straight out. And here they're throwing the projectile at an angle downwards. And so what we're going to do in this video is think about for each of these initial velocity vectors, what would the acceleration versus time, the velocity versus time, and the position versus time graphs look like in both the y and the x directions. So I encourage you to pause this video and think about it on your own or even take out some paper and try to solve it before I work through it. So let's first think about acceleration in the vertical dimension, acceleration in the y direction. We're assuming we're on Earth and we're going to ignore air resistance. We can assume we're in some type of a laboratory vacuum and this person had maybe an astronaut suit on even though they're on Earth. What would be the acceleration in the vertical direction? Well the acceleration due to gravity will be downwards, and it's going to be constant. We're going to assume constant acceleration. So the acceleration is going to look like this. And if the magnitude of the acceleration due to gravity is g, we could call this negative g to show that it is a downward acceleration. Once the projectile is let loose, that's the way it's going to be accelerated. Now what about in the x direction? Well if we assume no air resistance, then there's not going to be any acceleration or deceleration in the x direction. So it's just going to be, it's just going to stay right at zero and it's not going to change. And what I've just drawn here is going to be true for all three of these scenarios because the direction with which you throw it, that doesn't somehow affect the acceleration due to gravity once the ball is actually out of your hands. So now let's think about velocity. So what is going to be the velocity in the y direction for this first scenario? Well we could take our initial velocity vector that has this velocity at an angle and break it up into its y and x components. So this would be its y component. We just take the top part of this vector right over here, the head of it, and go to the left, and so that would be the magnitude of its y component, and then this would be the magnitude of its x component. So the y component, it starts positive, so it's like that, but remember our acceleration is a constant negative. So our velocity is going to decrease at a constant rate. So our velocity in this first scenario is going to look something, is going to look something like that. Now what about the velocity in the x direction? We see that it starts positive, so it's going to start positive, and if we're in a world with no air resistance, well then it's just going to stay positive. Notice we have zero acceleration, so our velocity is just going to stay positive. One of the things to really keep in mind when we start doing two-dimensional projectile motion like we're doing right over here is once you break down your vectors into x and y components, you can treat them completely independently. That something will decelerate in the y direction, but it doesn't mean that it's going to decelerate in the x direction. Now what would the velocities look like for this blue scenario? Well our velocity in our y direction, we start off with no velocity in our y direction so it's going to be right over here. But then we are going to be accelerated downward, so our velocity is going to get more and more and more negative as time passes. And notice the slope on these two lines are the same because the rate of acceleration is the same, even though you had a different starting point. Now what about the velocity in the x direction here? It looks like this x initial velocity is a little bit more than this one, so maybe it's a little bit higher, but it stays constant once again. Now let's look at this third scenario. In this third scenario, what is our y velocity, our initial y velocity? Well it would look something like that. And our initial x velocity would look something like that. If we were to break things down into their components. So our y velocity is starting negative, is starting negative, and then it's just going to get more and more negative once the individual lets go of the ball. Because you have that constant acceleration, that negative acceleration, so it's gonna look something like that. And what about in the x direction? Well looks like in the x direction right over here is very similar to that one, so it might look something like this. I'll draw it slightly higher just so you can see it, but once again the velocity x direction stays the same because in all three scenarios, you have zero acceleration in the x direction. Now last but not least let's think about position. So they all start in the exact same place at both the x and y dimension, but as we see, they all have different initial velocities, at least in the y dimension. So let's start with the salmon colored one. So the salmon colored one, it starts off with a some type of positive y position, maybe based on the height of where the individual's hand is. And then what's going to happen? Well it's going to have positive but decreasing velocity up until this point. At this point its velocity is zero. So its position is going to go up but at ever decreasing rates until you get right to that point right over there, and then we see the velocity starts becoming more and more and more and more negative. So it would look something like, something like that. Now what would be the x position of this first scenario? Well if we make this position right over here zero, then we would start our x position would start over here, and since we have a constant positive x velocity, our x position would just increase at a constant rate. It would do something like that. Now what about this blue scenario? Well this blue scenario, we are starting in the exact same place as in our pink scenario, and then our initial y velocity is zero, and then it just gets more and more and more and more negative. So it would look something, it would look something like this. Now what about the x position? Well our x position, we had a slightly higher velocity, at least the way that I drew it over here, so we our x position would increase at a constant rate and it would be a slightly higher constant rate. So it would have a slightly higher slope than we saw for the pink one. Now the yellow scenario, once again we're starting in the exact same place, and here we're already starting with a negative velocity and it's only gonna get more and more and more negative. So it's just gonna do something like this. It's gonna get more and more and more negative. It's a little bit hard to see, but it would do something like that. And if the in the x direction, our velocity is roughly the same as the blue scenario, then our x position over time for the yellow one is gonna look pretty pretty similar. So this is just a way to visualize how things would behave in terms of position, velocity, and acceleration in the y and x directions and to appreciate, one, how to draw and visualize these graphs and conceptualize them, but also to appreciate that you can treat, once you break your initial velocity vectors down, you can treat the different dimensions, the x and the y dimensions, independently.