Lab Report

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Problem Statement: How will the rate of the object change when dropped under water as opposed to in the air?

Hypothesis: Since the amount of drag depends on the surface area, desity, drag coefficient, and velocity, the different objects will have more drag when dropped underwater as opposed to when dropped in air. We also think because of the different surface areas, densities, drag coefficients, and velocities, the different objects will have different times to hit the ground.

 

Method and Materials: Our experiment can be done by following these given instructions.

1. Set the height for the objects to be dropped above and under water
2. Drop each object from the desired height three times
3. Record how long it takes for each object to fall
4. Use formula D/T=S to calculate speed of objects
5. Record data and compare and contrast results of in air and in water

These are the materials that we will use for this experiment.
1. Square rock
2. Full soda can
3. CD
4. Metal spike
5. Landing pad
6. Tape measurer
7. Stopwatch

Results:

Graph of averages


Times for the items dropped in water.

 

 

As explained in our analytic essay, our project has error. This is how you calculate exactly the percent amount of error for results.

1. We took the average of the times for our three trials by adding them together and dividing by three.

2.Then you take the time from trial 1 and subtract the average, the time from trial two and subtract the average, and the time from trial three and subtract the average.

3. You then add all those answers together and divide the answer of that by three

4. You then take the answer to that and divide by the average of the three trials

5. Multiply the answer by 100 and you have the percent error

This is the percent error for the times in water.

Full soda can:
18.97-18.443=.527
18.53-18.443=.0867
17.88-18.443=.564

(.527+.0867+.564)/3=.393

.393/18.443=0.021*100=2.1%
The percent error for the full soda can in water is 2.1%

CD:
27.72-27.563=.157
26.94-27.563=.623
28.03-27.563=0.467

(.157+.623+.467)/3=1.354

1.354/27.563=.049*100=4.9%
The percent error for the CD in water is 4.9%

Metal Spike:
3.41-3.55=.14
3.86-3.55=.31
3.38-3.55=.17

(.14+.31+.17)/3=.206

.206/3.55=.058*100=5.8%
The percent error for the metal spike in water is 5.8%

Rectangular Rock:
5.03-5.03666=.006
4.98-5.03666=.056
5.1-5.03666=.064

(.006+.056+.064)/3=.042

.042/5.03666=.0083*100=.83%
The percent error for the Rock in water is 0.83%

 

This is the percent error for the times in air.

Full soda can:
.84-.873=.03
.94-.873=.06666667
.84-.873=.03

(.03+.06666667+.03)/3=.04433/.873=.0509*100=5.09%

CD=2.04%

Metal Spike=4.6666667%

Rectangular Rock=2.13%

Conclusion: From our results, we can see that our hypothesis was mostly correct.In both water and in air, the CD took the longest time to hit the ground and the spike took the least amount of time. Roughly, the times in air were 4.8% of the times in water.