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 has roots  & 
Thus,
g(x) = (x-)(x-) - eq.1
We have,
P = 
Replacing g(x ) as in eq .1, we get
P = [ ] [ ] {ef(x-)} / { ef(x-) + ef(x-)}.dx - eq.2
We know,
[a] [b]f(x)dx = [a][b]f(a+b - x)dx
Applying the same in eq. 2
P = [ ][ ] {ef(-x)} / { ef(-x) + ef(-x)}.dx
As f(x) is even,
f(-x) = f(x),
P = [ ][ ] {ef(x-)} / { ef(x-) + ef(x-)}.dx eq.3
Adding eq.2 and 3,
the numerator n denominator get cancelled and we get,
2P = [ ][ ]1.dx
2P =  -
P = ( - )/2
Thus, the solution is B
as - = ( -b+D)/2a - ( -b - D)/2a
= 2D/2a
= D/a
Hope the solution satisfies you well..
Plz Rate Me..
If u have any query, nudge me!!
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x + 1/x is always more than equal to 2
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How many times are u gonna ask the same question??
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The hint is enough..Use your brain and not others' !!
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Thx everbody...
What would happen if the ball on the righmost side is of mass 2m...Then, will 2 balls lift or only the last ball ??
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I'm speaking exactly opposite of ur assumption..
I'm not releasing the 2 balls after a time interval.. I'm releasing the balls simultaneously..
Then even two balls will rise with momentum ( mv + mv)
and not 1 ball with a momentum ( m *2v)
Why???
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R is the range of the projectile..
It is the maximum horizontal displacement of the projectile.
When a projectile is launched, the moves a certain horizontal distance before it comes at the same initial height. This horizontal distance is called the range of the projectile.
But the R sinjan has used in his first post is just the horizontal displacement of the projectile at any time t.
Hope u understood it..If not, nudge me!!
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You are right but here x has three roots which are equal..
So, the three values are
x=2; x=2 and x=2...
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I've taked the initial height of the projectile to be y=2m. i've not taken the point A to be the origin.
Thats why, we can take the final height to be 1...
The equation is like this
yfinal = yinitial + uyt + 1/2 ayt2
Here y initial is 2m and not zero m( as the origin of frame of reference is 2 metres below A ...
and y final is 1 m...
Displacement in vertical direction is -1m no doubt buddy!!
See the figure for clarification.
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This requires Elliptic Integrals of Second Kind..It is Out of course!!
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Here, in the figure, we see 6 balls of similar masses attached.
Now, if we lift two balls from the left hand side, and leave them to fall down to strike the remaining balls, two balls from the right will lift up.
I'd like to know why always 2 balls would lift up?
Why cant one single ball lift up with an equal momentum, i.e.m(2v) and not (mv + mv).
Plz explain this...
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For the ball, the initial velocity, u = 7.8 m/s at an angle 30o to horizontal
Then,
Vx = ucos = 7.8*( 3/2) = 6.75m/s
Vy = usin - (1/2) gt2= 3.9 m/s - 4.9t2
X= ucost = 6.75t eq.1
Y = 2+ usint -(1/2)gt2 = 2 + 3.9t - 4.9t2
Putting Y = 1, we get the quadratic in t as
4.9t2 - 3.9t -1 = 0
t = {3.9 +/- (3.92 + 4*4.9)} / 2*4.9
t = (3.9 +/- 5.9)/9.8
Taking positive value,
t = 1sec
Puhtting t=1 in eq.1
X = 6.75*1
X=6.75
The distance between A and B is 6.75m
HERE's YOUR SOLUTION
PLZ RATE!!
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@1357924680
plz give the correct answer!!
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What is step??
and for the first part, do u mean
(b +  c) , ( a + c ),( a + b) are in HP???
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v1=(m1-em2)/(m1+m2)+(m2+em2)/(m1+m2) v2=(m2-em1)/(m1+m2)+(m1+em1)/(m1+m2)
Plz dont give direct formulae..
Derive them.. That would help us to solve other questions too!!
Thx
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@bhuwan
for the ans u gave for q.2, tell us what is
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See my solution...That would solve whatever mistake u did...
RATE MY SOLUTION Plzz
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The forum creator isn't even giving the right answer!!!
Lets skip the thread now!!
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My solution takes the direction of B the same as that of A!!
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