Video Analysis of moving objects

Have you ever wondered what the video analysis of a ball thrown through the air would be?
Prepare to be amazed, because here it is.

Here is our ball's track, the same video done twice:

These captions show the path of our ball over the course of its flight. One of the things that needs attention is that we have the meter stick, which gives a frame of reference for the flight of the ball of 1m.

This is the graph that shows both our change in position in the x direction (green), and in the y-direction (red). We can see how much distance it covers, how long it takes, and why.

When we ran our velocities, we found the following data:

This graph shows the velocities, and thus the accelerations of our graphs. In the x direction, we can see that the displacement is very low, and our graph peaks then seems to come back to almost the same as when it started. However in the y direction, our ball seems to increase its speed for the first few seconds, but then it comes to reaches a crescendo, the peak, and slows down until it hits zero, which is the point where it starts to descend.

A and B. 

Our acceleration in the y direction is 9.81 m/s^2 due to the force of gravity, and our acceleration in the x direction is 0m/s^2 as x-direction does not have an acceleration capacity, rather an ability to exist in that of constant velocity as the change in v evens out, making it constant velocity.

C & D. 

Our initial velocity in the x-direction is 5 m/s which remains constant the whole way through, but our initial velocity in the y-direction 3.09 m/s as that is the initial slope on the position vs time graph for the y direction.

E&F. 

The velocity at the top of our path in the x direction is 5 m/s, while the velocity at the top of our path in the y direction is 0 m/s as that is the point where it is about to turn around, thus reaching standstill for a moment.

G&H

The final velocity in the x direction remained the same, 5 m/s as it was travelling at a constant velocity, but had reached -9.65 on the last part of the graph.

                                                   I&J.

In order to discover the height of the ball, we can see where it was on our position graph, and it was 1.8m. The distance our ball traveled can found using x_final-x_initial=change in x. In this case, our change in x was 2-0.4=1.6 m.

K&L

It took 2 seconds for the ball to reach the top of its path (27.5s-25.5s), and it

Remained in the air (29.15-25.5) 3.65s. 

Conclusion:

1. Vertical accelerations are constant at 9.81 m/s^2 as a result of the force of gravity, but horizontal acceleration is 0 m/s^2, as it is travelling at constant velocity, not acceleration. I know this both through equations, and also the visible aspect of our data. 

2. For Vertical formula's, I can use the Δx = 1/2 aΔt² + ViΔt formula, but for the horizontal, the only formula we can use is v = Δxv/Δt, and let that solve it. In order to do so, I would solve for time through the change in x formula, and that input that into the velocity formula. These two formula's work together, not against each other. 

3. In order to solve for the displacement, I would use the Δx=xf-xi, and find it from there, but the vertical formula is a longer verison that has 0's in it to cancel it out. 

4. In order to solve for height, we would use the Δx = 1/2 aΔt² + ViΔt, input time, and then divide our x by two, as it would take equal amounts of time to fall to its peak, and to fall to the ground. 

5. At the top of its peak, there is only the horizontal velocity, as the vertical velocity is nonexistent.