Physics element in october sky

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SUMMARY

The discussion focuses on the physics elements presented in the film "October Sky," particularly the concepts of projectile motion and parabolic trajectories. A key scene involves Homer using a chart to demonstrate the rocket's landing location, applying the equation d = v0^2 * sin(2θ)/g to calculate distance, where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (32 feet per second squared or 9.8 meters per second squared). The film effectively illustrates the principles of conservation of energy and the significance of physics in achieving scientific goals.

PREREQUISITES
  • Understanding of projectile motion
  • Familiarity with parabolic equations
  • Knowledge of the conservation of energy principle
  • Basic grasp of gravitational acceleration (32 ft/s² or 9.8 m/s²)
NEXT STEPS
  • Study the physics of projectile motion in detail
  • Learn how to derive and apply the parabolic motion equation
  • Explore conservation of energy in practical experiments
  • Investigate real-world applications of physics in rocketry
USEFUL FOR

Students, educators, and anyone interested in the application of physics in real-life scenarios, particularly in the context of rocketry and project-based learning.

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physics element in "october sky"

hi, my physics teacher gave us a project, we watched "October Sky" in class, and the teacher wants us to do a two page report on the "physics element" of october sky, one of the most prominent moment with the most physics presented is when Homer trys to prove to his principal that the rocket didn't start the fire, and in fact the rocket landed elsewhere.. i can't understand the physics behind his explanation, and also don't understand his chart, he used parabola for finding where the rocket landed.. can someone tell me (or refer me) how do you suppose to find out how long a distance a rocket going in parabola has traveled?
 
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This scene is where Homer discusses "projectile motion." Have you done this yet? In the scene he refers to the acceleration of gravity as "32 feet per second squared." You might have heard of the metric equivalent "9.8 meters per second squared."

Anyway, I assume you have a textbook. Look up "projectile motion" and it will have plenty of parabolas there.
 


There are several physics elements present in "October Sky" that contribute to the overall storyline and theme of the movie. One of the most prominent is the use of parabolic motion and projectile motion in the rocket launches.

In the scene where Homer is trying to prove the rocket's landing location to his principal, he uses a chart and explains how the rocket's trajectory would follow a parabolic path. This is due to the force of gravity acting on the rocket as it is launched into the air. The rocket's velocity and angle of launch determine the shape and distance of the parabola.

To calculate the distance traveled by the rocket, you can use the equation d = v0^2 * sin(2θ)/g, where v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. This equation takes into account the horizontal and vertical components of the rocket's motion.

In addition to parabolic motion, the movie also showcases the principle of conservation of energy. The boys have to carefully calculate the amount of fuel and the angle of the launch to ensure that the rocket reaches the desired height. This shows how understanding and applying physics concepts can lead to successful experiments and achievements.

Overall, the physics element in "October Sky" adds a layer of realism and depth to the story, as well as highlighting the importance of science and technology in our daily lives. It also serves as a reminder of the power of determination and perseverance in pursuing our dreams and goals.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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