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 blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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 Help with some simple physics problems 
#3657187 - 01/21/05 02:11 AM (19 years, 2 months ago) |
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Im trying to figure out some simple physics problems to help someone, but unsucessfully. The first one is:
The corners of an equilateral triangle lie on a circle of radius 2.0 m. Find the length of a side of the triangle.
I really don't know how to approach that one. Its been a while.
The next one is:
You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of 14 m with the same initial speed of 12 m/s, but in different directions. You through your snowball downward, at 40? below the horizontal; your friend throws her snowball upward, at 40? above the horizontal. What is the speed of each ball when it is 5.0 m above the ground?
Obviously they will both have the same speed...but I can't get the right answer for some reason.
To find the time it takes to reach 5 meters, I used the equation:
y=(Vo sin theta)t-.5gt^2, and then used the quadratic equation. Am I wrong in assuming this would be the proper way to find the time?
Next I simply calculated the velocity with respect to x using Vx=Vo Cos theta
and for the y I used Vy=Vo sin theta - gt.
Am I approaching this wrong, or I am simply making a computational error?
Any help would be greatly appreciated.
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RandalFlagg
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Registered: 06/15/02
Posts: 15,608
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I think I know how to do this one.
An equilateral triangle is one where all of the sides are the same length. This means that right smack in the middle of that triangle is the same middle point as the circle. Now, look at that triangle again. Draw a straight line from the middle point to one of the triangle's corners(on the edge of the circle). Draw another straight line from the middle point to another of the triangle's corners. Because you started from the middle point and ended up at the edge of the circle, and your radius is 2m, you have just made a 2m, 2m, X(X being the unkown quantity of the side of the equilateral triangle) triangle.
An isosceles triangle (a triangle with two equal sides) has inside angles of 40 deg., 40 deg., and 100 deg. So now look at the isosceles triangle you just made. Split it down the middle so you now have two right triangles of 2m(the hypotenuse), 1/2(X), and another unknown. Because you have the hypotenuse and the angle of a right triangle, you can determined the sine. With the hyptoenuse and the sine, you can determine the length of the base of this right triangle. Take that number and multiply by two and you have length of one of the sides of the original equilateral triangle.
The anwser to problem is 3.064177m. That is how long the sides of the equilateral triangle are.
Edited by RandalFlagg (01/21/05 04:40 AM)
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 Anno
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Registered: 06/17/99
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Last seen: 4 days, 19 hours
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>The corners of an equilateral triangle lie on a circle of radius 2.0 m. Find the length of a
>side of the triangle.
Obviously , as the triangle is symmetric, its center of gravity lies in the middle of it, thus in the middle of the circle.
Now, the center of gravity is on 1/3 of the triangle height.
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RandalFlagg
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As for that physics problem....I could whip out my old books and look up some equations and probably figure out that problem. But, I promised myself I would never do a fucking physics problem again!! Fuck that shit. I still get nightmares about sitting in physics class in college and doing complicated problems, running out of time, and failing.
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: RandalFlagg]
#3657353 - 01/21/05 04:08 AM (19 years, 2 months ago) |
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Quote:
RandalFlagg said:
I think I know how to do this one.
An equilateral triangle is one where all of the sides are the same length. This means that right smack in the middle of that triangle is the same middle point as the circle. Now, look at that triangle again. Draw a straight line from the middle point to one of the triangle's corners(on the edge of the circle). Take another straight line from the middle point to another of the triangle's corners. Because you started from the middle point and ended up at the edge of the circle, and your radius is 2m, you have just made a 2m, 2m, X(X being the unkown quantity of the side of the equilateral triangle) triangle.
An isosceles triangle (a triangle with two equal sides) has inside angles of 40 deg., 40 deg., and 100 deg. So now look at your triangle again. Split it down the middle so you now have two right triangles of 2m(the hypotenuse), 1/2(X), and another unknown. Because you have the hypotenuse and the angle of a right triangle, you can determined the sine. With the hyptoenuse and the sine, you can determine the length of the base of this right triangle. Take that number and multiply by two and you have length of one of the sides of the original equilateral triangle.
The anwser to problem is 3.064177m. That is how long the sides of the equilateral triangle are.
Hmm, that answer may be correct, but the program is saying its incorrect, even with a 2% tolerance. 
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RandalFlagg
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Registered: 06/15/02
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Really? If you recreate everything I said on paper, it seems as if it would be mathematically correct.
A right triangle with a 2m hypotenuse and a 40 degree angle would
have to be a 2, 1.53208888624(base), 1.28557521937(side) triangle.
And 1.532 * 2 = 3.064
Maybe you gave the wrong measurement for the radius? Did you get the diameter and radius mixed up?
Edited by RandalFlagg (01/21/05 04:32 AM)
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: RandalFlagg]
#3657386 - 01/21/05 04:32 AM (19 years, 2 months ago) |
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Quote:
RandalFlagg said:
Really? If you recreate everything I said on paper, it seems as if it would be mathematically correct.
A right triangle with a 2m hypotenuse and a 40 degree angle would
have to be a 2, 1.53208888624(base), 1.28557521937(side) triangle.
And 1.532 * 2 = 3.064
Yeah, we tried it twice but with no success, and I don't see any flaws in your reasoning stupid physics
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RandalFlagg
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Registered: 06/15/02
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WTF!! Have I become an idiot or something? This has to be right. Maybe that program is wrong. I don't see how I could be wrong. I have worked it out on paper and everything.
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: RandalFlagg]
#3657400 - 01/21/05 04:39 AM (19 years, 2 months ago) |
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Quote:
RandalFlagg said:
WTF!! Have I become an idiot or something? This has to be right. Maybe that program is wrong. I don't see how I could be wrong. I have worked it out on paper and everything.
The program could be wrong, since it was made by the instructor, however I have never encountered it being wrong.
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RandalFlagg
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Undeniable Mathematical proof!!
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: RandalFlagg]
#3657435 - 01/21/05 05:11 AM (19 years, 2 months ago) |
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Quote:
RandalFlagg said:
Undeniable Mathematical proof!!
Looks sound to me , either my prof is wrong, or we are missing something here. 
If my prof is wrong, I should totally get extra credit, despite me not being the one to solve the problem.
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Anno
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Registered: 06/17/99
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See my solution above.
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Anno
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Posts: 24,166
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Re: Help with some simple physics problems [Re: RandalFlagg]
#3657472 - 01/21/05 05:44 AM (19 years, 2 months ago) |
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> 40 degree angle
why 40?
If you split a 60? angle in half, you get 30?.....
The solution is 3.46410162 btw
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: Anno]
#3657497 - 01/21/05 06:20 AM (19 years, 2 months ago) |
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Quote:
Anno said:
> 40 degree angle
why 40?
If you split a 60? angle in half, you get 30?.....
The solution is 3.46410162 btw
You sir are absolutely correct. Thankyou very much.
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Anno
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Since you want the velocity at 5m, you could also stand on a 9 m high roof and seek the velocity at the ground level.
The initial vertical velocity is the same both times, once down, once up.
Important are only the energies, the potential energy and the kinetic energy.
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: Anno]
#3657522 - 01/21/05 07:00 AM (19 years, 2 months ago) |
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Quote:
Anno said:
Since you want the velocity at 5m, you could also stand on a 9 m high roof and seek the velocity at the ground level.
The initial vertical velocity is the same both times, once down, once up.
Important are only the energies, the potential energy and the kinetic energy.
Hmm, I'm not sure, but I think a velocity of 73k is a little too high? Or did I miss something. I can't believe I forgot about solving these problems using energy rather than vectors.
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RandalFlagg
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Re: Help with some simple physics problems [Re: Anno]
#3657525 - 01/21/05 07:05 AM (19 years, 2 months ago) |
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Shit! Shit! Shit!
I am wrong.
My methodology was sound, but I fucked up with the 40 degree assertion. I was looking at a diagram of an isosceles triangle and I saw that it was a 40 deg., 40 deg, 100 deg. triangle and my stupid brain assumed that that was the degree law of those particular triangles. As I thought about it more I realized that the base could be any length....so the only law of isosceles triangles is that two of the angles are equal....I am an idiot!
Ok...let's use my methodology to solve it correctly this time...
On step 2 of my picture I have divided the equilateral triangle into one isosceles. If you put another line fron the mid. point to the last corner of the equilateral triangle, you will have seperated it into three equal isosceles triangles. If you look at that mid point where the tops of all three triangles intersect, it is 360 degrees all the way around. This means that the tops of all three of those isosceles triangles are 120 deg. a piece. Which means that instead of a 40, deg. 40 deg., 100 deg. triangle in step 3 of my diagram, it is 30 deg., 30 deg., and 120 deg.
Then if you divide the isosceles into a right triangle(like in my step 4) and do the trigonometry, you come up with the base of the divided right triangle equaling 1.73. Take that times two and you get 3.46
Sorry for fucking your answer up Blacksabbathrulez. I kind of had it right.
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Anno
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Registered: 06/17/99
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> I think a velocity of 73k is a little too high?
It is 13.907 m/s
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Re: Help with some simple physics problems [Re: Anno]
#3657533 - 01/21/05 07:12 AM (19 years, 2 months ago) |
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Quote:
Anno said:
> I think a velocity of 73k is a little too high?
It is 13.907 m/s
Thanks again 
PS. awesome avatar.
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blacksabbathrulz
Registered: 05/22/02
Posts: 2,511
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Quote:
blacksabbathrulz said:
Quote:
Anno said:
> I think a velocity of 73k is a little too high?
It is 13.907 m/s
Thanks again
PS. awesome avatar.
Hmm, I just checked and that was an answer I already gave. That's the answer I initially got, and I'm almost positive its correct. I think the teacher may be in error in this case.
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