
minusrestraint
i came to BRINGTHE PAIN. andthe punch :D.
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physics help
#1256587  01/28/03 07:48 PM (13 years, 8 months ago) 


i posted this in OTD cuz i forgot what a waste of space it could. i hope someone can help me.
ok, im doing my physics 106 homework. i have this problem, and i cant figure it out. i know its an easy 2D kinematics equation, but christ idk. ill type out the problem below.
An object is released from rest at an unknown height h above the ground. One second later a second object is released from rest at the same height. When the first object strikes the ground, the second is 20 m above the ground. What is the initial height h?
could you guys help any? the variables to use are initial velocity, final velocity, time, accelleration, and distance (height).
so far in this class, she's only given us four equations to choose from..: 1 ) V(f)=at+V(i) 2 ) V= change in distance/time 3 ) d=(1/2)(a)(t squared) + V(i) (t) 4 ) 2ad = v(f)squared + v(i)squared
i would REALLY appreciate some help on this, its been stumping me for two days and its due tomorrow
 "is there a doctor in the house?
we like fuck that, nut sacks in yo mouth
lemme show you what a thug about
we can talk or we can slug it out"
cisco kid

zeta
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Personally I would just use h = 1/2 a*t^2 First object: h = 1/2 a*t^2 Second object: h  20 = 1/2 a*(t1)^2 Rearrange both equations to make t (or t^2) the subject and equate them to each other, solve for h I hope you understand this.. if I can be bothered working it out then I'll post the answer

Jackal
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The two balls take the same journey time to hit the ground, so will hit the ground 1 second apart. When the first ball hits the ground the second still has 20m to travel, so using eqn 3 with a=10, d=20 and t=1 we can say that the second ball is travelling at V=15 m/s at the moment the first ball hits the ground.
But the second ball was dropped from rest some time previously. Using eqn 1 with V(i)=0, a=10 (again) and V(f)=15 we see that the second ball was dropped 15/10=1.5 seconds ago. So the total journey time for a ball to hit the ground is 2.5 seconds.
Use eqn 3 again with initial velocity zero to show that the height must be 1/2x10x2.5x2.5=31.25m

minusrestraint
i came to BRINGTHE PAIN. andthe punch :D.
Registered: 09/04/02
Posts: 1,093
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Last seen: 11 years, 8 months

Re: physics help [Re: Jackal]
#1257574  01/29/03 07:03 AM (13 years, 8 months ago) 


wow jackal. ive gotten like.....6 answers reponses...and yours was the first i understood. 5 shrooms. i award you.
too bad i didnt see the post until i got back from that class.
 "is there a doctor in the house?
we like fuck that, nut sacks in yo mouth
lemme show you what a thug about
we can talk or we can slug it out"
cisco kid

Baby_Hitler
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Re: physics help [Re: Jackal]
#1258001  01/29/03 09:38 AM (13 years, 8 months ago) 


Hmm... I got 31.61m


Earth_Droid
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Do I get some mushrooms for recommending you post it at the chill room lol

recalcitrant
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ya, i have a math question too.
you have a circular cone with a radius of 1 and a height of 3. there is a cube inside the cone with one face flush with the face of the cone. the opposite four points of the cube are touching the inside of the cone. how do you find out the dimensions of the cube?

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zeta
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What value of acceleration due to gravity were you using?

Baby_Hitler
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Re: physics help [Re: zeta]
#1259484  01/29/03 05:20 PM (13 years, 8 months ago) 


9.8m/s^2


Anonymous

Post deleted by Moe Howard [Re: Baby_Hitler]
#1259485  01/29/03 05:22 PM (13 years, 8 months ago) 



minusrestraint
i came to BRINGTHE PAIN. andthe punch :D.
Registered: 09/04/02
Posts: 1,093
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Last seen: 11 years, 8 months

Re: physics help [Re: ]
#1259538  01/29/03 05:45 PM (13 years, 8 months ago) 


you guys are so sexy. thanks for the help sexies.
 "is there a doctor in the house?
we like fuck that, nut sacks in yo mouth
lemme show you what a thug about
we can talk or we can slug it out"
cisco kid

Jackal
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OK here goes:
Place the cone with its vertex (the pointy part) up, and the face of the cone against the xyplane. The cube sits on the xyplane, inside the cone, you described.
Next, take a vertical slice through the assembly, so that it passes through four of the eight vertices of the cube, as well as the vertex of the cone.
Here's what that slice looks like (you'll have to use a little imagination, due to the limitations of ASCII graphics): ................_C ^  /\  /  \  3s /  \  /  \  /+++\...........+B /   \  /   \  /   \ s /   \  ++.......A 1sqrt(2)s/2
Now, the bottom line is a diameter of the circle forming the base of the cone, and so has length 2. The side of the cube is s, so that's also the height of the (rectangular) slice through the cube. The horizontal sides of this slice are diagonals of the square top & bottom, and so they have length sqrt(2)s. The cone has height 3, so that's the height of the triangular slice you see. Since the cube must rest in the center of the circular base, the two segments on either side of the rectangle, inside the triangle, are equal and thus of length (2sqrt(2)s)/2, or 1(sqrt(2)s/2). Also, the two parts of the rectangle to the right and left of the center line I've drawn are also equal to each other, so they must be of length sqrt(2)s/2.
Now, the two right triangles, one on the right half above the rectangle, and one to the right of the rectangle, having (respectively) parallel sides to each other, must be similar. Thus, their sides' lengths are in constant proportion to one another, and we can write down what those values are, in terms of the unknown side s of the cube:
Upper triangle: 3s, sqrt(2)s/2, and something ( diagonal ) Lower triangle: s, 1sqrt(2)s/2, and something ( diagonal )
Let's take the proportions of corresponding sides and write an equation expressing the fact that those proportions are equal:
(3s)/s = (sqrt(2)s/2)/(1sqrt(2)s/2)
Multiply through by the product of denominators:
(3s) (1sqrt(2)s/2) = (sqrt(2)s/2)s
and do the arithmetic:
3s  3sqrt(2)s/2 + sqrt(2)s^2/2 = sqrt(2)s^2/2
Now, bring all the terms involving s to the right and combine:
3 = s + 3 sqrt(2)s/2  sqrt(2)s^2/2 + sqrt(2)s^2/2
3 = (1 + 3 sqrt(2)/2) s
so, we find
s = 3/(1 + 3 sqrt(2)/2) = 6/(2 + 3 sqrt(2))
That fraction can be simplified somewhat by multiplying numerator and denominator by (3 sqrt(2)2):
s = 6 ( 3 sqrt(2)  2) / (14) = 3(3 sqrt(2)  2)/7
So, s is approximately 0.961 units.
That is, if I've done the algebra and arithmetic correctly.

Earth_Droid
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Re: physics help [Re: Jackal]
#1260374  01/30/03 03:28 AM (13 years, 8 months ago) 


Jackal is math expert extrordinaire!
I will know where to go when I have a math question.

recalcitrant
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Re: physics help [Re: Jackal]
#1260906  01/30/03 07:56 AM (13 years, 8 months ago) 


oic. ur ascii was a bit buggered tho

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Jackal
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I know, the BB kept taking the spaces out

