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Deceleration's 100 math problems thread!; Absolute torture...
Topic Started: Aug 4 2011, 04:22 AM (5,612 Views)
Deceleration
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Coreling Conquerer

I had one about how the sun and the moon appear about the same size from Earth, and finding a distance. But I forgot.
 
Camo5
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Resident BorgCube
I have one!

A. find the center of balance between two objects attached to either end of a 1.7m long barbell weighing 20kg and 35kg respectively.

B. Then find the center of balance by including the bar's weight of 8kg.
 
Brayzure
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The Analytic Commander

I have the answer for the scrapped answer!

Barring air resistance and gravitational anomalies, the projectile will go between 7.27368 km and 7.2740 km (my calculator gave me both when I did this one step).

Here's my process.

I created parametric equations to model the Earth and the path of the artillery shot.
Then, I removed the parameter to yield regular functions.
I found the intersection of them. (7274.0207,-4.147896)
The first point proved problematic, so I used the second one to solve for the y-parameter with the equation for Earth.
In Radians, I got an angle of .00114... for the angle between the top, and where the shot impacted.
I then multiplied it by 6378100 (the radius of the Earth in meters) to get the arc length in meters: 7273.832678 m, or 7.2738 km.
I followed a similar process with degrees and got an answer about a third of a meter larger.
 
Deceleration
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Coreling Conquerer

Huh, that's odd, I got an answer about 50 times larger than that... I did it with 1000 m/s as initial velocity and got 181 km or thereabouts. I discovered my mistake and decided to give up, though it only took about 10 minutes of head scratching. Maybe I'll do it properly later in pre-calculus.
 
Brayzure
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The Analytic Commander

Well, what did you do? Have you even learned parametric equations?
 
Deceleration
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Coreling Conquerer

I don't know, I used equations of motion. I split the initial velocity into two vectors, up and sideways, and solved for the time for the shell to hit the water using the initial vertical velocity. Then I solved for the distance traveled during that time using the horizontal vector, and that was my answer. The shell should really travel farther than 7.4 kilometers, as an initial horizontal velocity of about 1,700 meters per second would cover that in about 4 and a half seconds, way too short for the flight of the shell.
 
Brayzure
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The Analytic Commander

That might be true at a steeper angle. Unless I'm mistaken, a 30 degree angle forces any projectile to hit the ground/water pretty fast.
 
Deceleration
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Coreling Conquerer

Not that fast... 45 degrees provides the farthest distance traveled for a projectile. 60 and 30 degrees have the same distance, it's just that 30 degrees gets there faster and 60 degrees goes higher. I urge you to try again, Synergy. I am completely certain you are wrong.
 
Brayzure
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The Analytic Commander

Meh, as you wish.
 
Deceleration
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Coreling Conquerer

I worked it out! You were very much incorrect. Would you like me to explain how? Or would you like to post another guess?
 
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