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Galileo and the free fall of bodies

Exercise 1

Is it possible to represent the data (y, t) on a graph? Do it.

Exercise 2

With the data obtained to calculate the ball speed versus time for each interval. Observe that the average rate of displacement is increased with respect to time:

v (t) = increase and / t increase

Keep in mind that what calculáis represents the average velocity in an interval. This is an approximation to what would be the right thing: take the instantaneous velocity of the ball at each point. Remember that this is a MRUA.

Since it is a movement speed MRUA constantly changes so we will calculate the velocity at each point raised. With the results we obtain the average speed with which the ball travels.

V = e / t

Position 0 → 0 / 0 = 0 m / s

Position 1 → (0 Position - Position 1) → 0025 / 0.08 = 0.3125 m / s

Position 2 → ( Position 1 - Position 2) → (0.12-0.025) / (0.16-0.08) = 1. 1875 m / s

Position 3 → ( Position 2 - Position 3) → (0.27-0.12) / (0.24-0.16) = 1,875 m / s

Position 4 → ( Position 3 - Position 4) → (0.49-0.27) / (0.32-0.24) = 2.75 m / s

Position 5 → ( Position 4 - Position 5) → (0.78-0.49) / (0.4-0.32) = 3,625 m / s

Position 6 → ( Position 5 - Position 6) → (1.13- 0.78) / (0.48-0.4) = 4,375 m / s

Exercise 3

With the data graph speed for each segment in terms of time and qualitative analysis of this graph. What can you say about the type of motion that describes the steel ball in his fall? Do you agree that observation with your expectations?

This graph represents the speed for each segment, ie how to increase the time and space also increases, which means that the mobile is moving as time passes. More specifically, what is happening is that the ball is moving (toward the ground) as it passes the time (obviously), that is, is falling with acceleration. It can be seen that as time passes, the graph is increasing tilt, which means that the rate increases. This is because it is uniformly accelerated rectilinear motion (MRUA), in this case, besides MRUA is a free fall, where the acceleration is known: gravity (9.8 m / s ²) .

This observation is consistent with our expectations, since the slope of the graph is higher compared with previous estimates gradually seen an increase in speed over time. We expect the ball to fall down (of course), it starts with an initial velocity of 0m/sy that was accelerating as time went on. That means more space that ran faster than anything else is released.

Exercise 4

A graph constructed from v (t) determines the value acceleration of gravity, g. Compare the value of g obtained with the familiar.

The height would be to use a spreadsheet such as Excel or Derive to list represent data and graphs. Then enough to copy the image and include it in the entry. If someone does not know how to do it, we will be happy to explain.

g = Δv / Dt = 9.8 m/s2

Our result was 10.1 m / s 2 , as you can see the margin error made with respect to the gravity of the earth that is 9.8 m / s 2 is just 0.3 m / s ^{2 .}

Exercise 5

If there is discrepancy between the theoretical model and obtained experimentally, detect and analyze possible sources of error. The theoretical model, ie what would have been obtained theoretically, it can be developed using the kinematic equations for free fall: h = v = gt 1/2gt2 (considering g = 9.8 m/s2) and represents the vt graph for the previous time values.

Usually, when doing an experiment of this magnitude, it is obvious that we always have physical errors: friction, calculation errors ... For this reason we always try to minimize these possible errors and obtain a more accurate result.

All data collected above are EXPERIMENTAL :

now show the same way THEORETICAL through the kinematic equations:

Position 2 -> We have chosen this position for his data, as an example, can obviously elegir cualquier posición.

h = 0,12 m
t = 0.16 s
v = 1,1875m/s

h = 1/2gt2

h = ½ · 9.8 · 0.16 ²

= 9,8 · 0.0256 / 2

= 0.25088 / 2 = 1.12

Thus we see that the height theoretical and experimental match.

v = gt

v = 0.48 = 9.8 ° 4.7 m / s

Thus once again to match the same way speed theoretical and experimental.

1.) 0,025 = 4,9 t ² → t = 0,07 s ; v = 0,69 m/s --> a = 0,69/0,07 = 9,8 m/s ²

2.) 0,12 = 4,9 t ² → t = 0,16 s ; v = 1,57 m/s --> a = 1,57 - 0,69/0,16 - 0,07 = 9 ,8 m/s ²

3.) 0,27 = 4,9 t ² → t = 0,23 s ; v = 2,25 m/s --> a = 2,25 - 1,57/0,23 - 0,16 = 9,8 m/s ²

4.) 0,49 = 4,9 t ² → t = 0,32 s ; v = 3,14 m/s --> a =3,14 - 2,25/0,32 - 0,23 = 9,8 m/s ²

5.) 0,78 = 4,9 t ² → t = 0,4 s ; v = 3,92 m/s --> a = 3,92 - 3,14/0,4 - 0,32 = 9,8 m/s2

6.) 1,13 = 4,9t ² → t = 0,48 s ; v = 4,7 m/s --> a = 4.7 to 3.92 / 0.48 to 0.4 = 9.8 m/s2

Exercise 6

One more thing: since we are immersed in the topic of work and energy, could you calculate the speed of the ball in point 6 by Theorem Conservation of energy?. Compare the data with that obtained using the kinematic equations for free fall motion: v = gt (taking g = 9.8 m/s2).

conservation law -> energy is neither created nor destroyed -> transforms.

1. The potential energy is greater the more up there. In our experiment the potential energy is greatest at the beginning of the experiment.

2. The kinetic energy increases with speed takes the object that suffers, so it will grow more space covered .

Potential Energy + Kinetic Energy = Total Energy

Potential Energy = m · g • h

kinetic energy = ½ ° m · v ²

(have to remove the kinetic energy versus time)

m · g • h = ½ ° m · v ² speed of the ball in the middle.

By equating the two equations, this allowing us to be out the ball speed at the midpoint and at this precise moment that both energies are equal.

Epotencial = ECinetica
m · g • h = ½ ° m · v
2 v = √ (2GH)
v = 4.704 m / s

The results are smooth in both equations as in the above charts. The error is minimal, which can be influenced by factors such as air friction or form of the ball.