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Offlineminusrestraint
i came to BRINGTHE PAIN. andthe punch :D.

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physics help
    #1256587 - 01/28/03 07:48 PM (13 years, 10 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 2-D 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


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Invisiblezeta
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Registered: 05/25/02
Posts: 3,972
Re: physics help [Re: minusrestraint]
    #1256942 - 01/28/03 10:31 PM (13 years, 10 months ago)

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*(t-1)^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


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OfflineJackal
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Re: physics help [Re: minusrestraint]
    #1257529 - 01/29/03 06:33 AM (13 years, 10 months ago)

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


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Offlineminusrestraint
i came to BRINGTHE PAIN. andthe punch :D.

Registered: 09/04/02
Posts: 1,093
Loc: a star but i see you fine
Last seen: 11 years, 10 months
Re: physics help [Re: Jackal]
    #1257574 - 01/29/03 07:03 AM (13 years, 10 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.  :frown:  :frown:


--------------------
"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


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OfflineBaby_Hitler
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Re: physics help [Re: Jackal]
    #1258001 - 01/29/03 09:38 AM (13 years, 10 months ago)

Hmm... I got 31.61m


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OfflineEarth_Droid
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Re: physics help [Re: minusrestraint]
    #1258033 - 01/29/03 09:51 AM (13 years, 10 months ago)

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


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Offlinerecalcitrant
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Registered: 04/20/02
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Re: physics help [Re: minusrestraint]
    #1258602 - 01/29/03 12:54 PM (13 years, 10 months ago)

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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Invisiblezeta
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Re: physics help [Re: Baby_Hitler]
    #1259261 - 01/29/03 04:09 PM (13 years, 10 months ago)

What value of acceleration due to gravity were you using?


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OfflineBaby_Hitler
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Re: physics help [Re: zeta]
    #1259484 - 01/29/03 05:20 PM (13 years, 10 months ago)

9.8m/s^2


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Anonymous

Post deleted by Moe Howard [Re: Baby_Hitler]
    #1259485 - 01/29/03 05:22 PM (13 years, 10 months ago)



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Offlineminusrestraint
i came to BRINGTHE PAIN. andthe punch :D.

Registered: 09/04/02
Posts: 1,093
Loc: a star but i see you fine
Last seen: 11 years, 10 months
Re: physics help [Re: ]
    #1259538 - 01/29/03 05:45 PM (13 years, 10 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


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OfflineJackal
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Re: physics help [Re: recalcitrant]
    #1260104 - 01/29/03 10:56 PM (13 years, 10 months ago)

OK here goes:

Place the cone with its vertex (the pointy part) up, and the face of the
cone against the xy-plane. The cube sits on the xy-plane, 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
                      ^                |
                    /|\              |
                    / | \              | 3-s
                  /  |  \            |
                  /  |  \            |
                /+---+---+\...........+B
                / |      | \          |
              /  |      |  \        |
              /  |      |  \        |s
            /    |      |    \      |
            -----+-------+-----.......|A
                                1-sqrt(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 (2-sqrt(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: 3-s, sqrt(2)s/2, and something ( diagonal )
Lower triangle: s, 1-sqrt(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:

(3-s)/s = (sqrt(2)s/2)/(1-sqrt(2)s/2)

Multiply through by the product of denominators:

(3-s) (1-sqrt(2)s/2) = (sqrt(2)s/2)s

and do the arithmetic:

3-s - 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.

:grin:


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OfflineEarth_Droid
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Registered: 04/19/02
Posts: 5,240
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Re: physics help [Re: Jackal]
    #1260374 - 01/30/03 03:28 AM (13 years, 10 months ago)

Jackal is math expert extrordinaire!  :grin:

I will know where to go when I have a math question.   


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Offlinerecalcitrant
My Own God

Registered: 04/20/02
Posts: 2,927
Loc: Canada West
Last seen: 6 months, 23 days
Re: physics help [Re: Jackal]
    #1260906 - 01/30/03 07:56 AM (13 years, 10 months ago)

oic. ur ascii was a bit buggered tho


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OfflineJackal
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Re: physics help [Re: recalcitrant]
    #1260997 - 01/30/03 08:31 AM (13 years, 10 months ago)

I know, the BB kept taking the spaces out :frown:


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