Rocket post

In this challenge, we were attempting to solve for how far our ball would go, after having solved for velocity at the 30& and 50 degree angles in order to solve for the 40 degree,
In order to do this, we collected the following data:

This leads to our estimates of



This leads to the math of

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. 

Cart and Ball post

UFPM Challenge

A modified Atwood machine:

[Image upload of diagram]

In this lab, we were attempting to calculate two variables. The constant velocity of our moving cart, and the rate at which our cart on the track was accelerating due to the hanging weight which was attached.

Our cart’s track had no angle, so all we needed to find was the rate at which our weight (50g, 0.5n) was causing it to accelerate.

Thus, the following force diagram was created.

Please note the lack of friction, thus causing our diagram to be Unbalanced.

Thus, our diagram becomes an unbalanced force particle model. In order to solve for such a situation, we use the a(acceleration)=f_net/m(kg)

Thus, our following formula became a=0.5/.6394

As a result, a=.78 m/s²

We measured how far our change in x was, we discovered that it was 90 cm which equals .9m. When we input our known variables into the Change in X=vit+ at².

The final formula to solve for time is .9=.78t². As a result, our time was 1.15 s.

Part 2 was testing the velocity of our cart on the ground, and thus we solved for the velocity of our cart, finding that it was 0.3 m/s. Thus we multiplied 0.3*1.15s ending up with 34.5 centimeters away.

In order to test our conclusion, we let our ball drop when our cart had reached 34.5 cm’s away from the meeting point.

Once we tested our predicament, it was found to be correct.

UFPM Blog

Unit Summary UFPM

This unit covered a vast amount of information in regards to unbalanced forces but was also really exciting. Examples of such forces are all balls in the air, as only gravity acts upon them, all accelerating objects, and anything really that is not maintaining a constant velocity.

The formula’s used were:

1.       vf = at + vi

2.       vf² = vi² + 2aΔx

3.       a = F_net/mass (in kg)

4.       Δx = 1/2 aΔt² + ViΔt

5.       a = Δv/Δt

We started this unit exploring what happens when there is more force going one way, than holding it back, by looking at what happened if friction, or air resistance were not factored in, or if they were, that they were not equal in force.

Examples of such scenarios are cars that are speeding up or slowing down. Part of this was learning how to solve for the frictional force, as shown below.

If a car that weighs 240 kg with an F_push  of 300n, and and F_F of 80, the following diagram looks like:

In order to solve for acceleration, we must use the a=fnet/m formula as shown below. 

We have solved for acceleration, it is .91m/s^2

If we know that a parachutist covered 1500 meters before expanding his parachute, from rest, how long did it take him?

                     It took him 17.3 seconds to fall 1500 meters.

Now, what is his final velocity right before he opens his parachute?

Thus it is 173m/s before he pulls his parachute cord open. 

This is the summary of our blog, if you have questions, please feel free to comment below.