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3 Jul 2009 10:51:12 IST

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a and b r moving in uniform gravitational field where the acceleration due to gravity is g initiall

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a and b r moving in uniform gravitational field where the acceleration due to gravity is g initially the partices located at o move with velocity v1 and v2 horizontally in opp direction at what time from this will they move mutually perpendicular

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NugoRama

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3 Jul 2009 11:33:23 IST

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Nice question ....any takers ?

Mr.Nasty ® Retired from Action.

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3 Jul 2009 11:35:56 IST

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edited

NugoRama

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3 Jul 2009 11:46:02 IST

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Umm ..wait nasty ...they wont be making 45 degrees each ...check that again..

~Áß???????

Blazing goIITian

Joined: 31 Aug 2008

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3 Jul 2009 12:08:22 IST

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gud question..

see the figure ......here the 2 particles hv been projected inopposite direction..

let particle 1 be..(blue path)..

now....suppose after time t ..the particles ..are at 90 deg to each other.

and at that point particle 1 makes angle

and particle 2 makes angle

so now .....for ..particle 1 ..

and for particle 2 ..now angle

so .............hence..

hence

so .....

i think i got it right..

BEST OF LUCK..(sorry for such big figure..)

~Áß???????

Blazing goIITian

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3 Jul 2009 12:11:45 IST

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hey....meher is ma answer correct???

meher sehgal

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3 Jul 2009 12:29:57 IST

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I am not sure as you seem to have taken tan theeta 1 taken by you will be vertical dist/hor distance. Theeta 1 at the position is angle which particle is making with the traject. of mtion at the point. Pl correct me if wrong.

~Áß???????

Blazing goIITian

Joined: 31 Aug 2008

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3 Jul 2009 12:36:20 IST

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hey....see ..

...........as u can see the figure.

i know that ...tan(theta 1 ) i hv taken from horizontal line to ..trajectory..

but if

than automatically angle between the projectiles path will be 90 deg ..( 180 - 90 )

hence thiscondition will work..!(U CHK THE ANSWER FRM BOOK NAA....)

BEST OF LUCK

~Áß???????

Blazing goIITian

Joined: 31 Aug 2008

Posts: 982

3 Jul 2009 13:06:47 IST

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hey...

look at this method also.

its more correct...!

here at the figure ..

let the vel of particle = V at position hen they are at 90 deg..(for particle 1 )

so ..now v1 is horizontal component..

hence ..

and now vertcal component..is ..

hence

and similarly

hence now

so ...

this answer is ......proposed by my friend rohit...!(balganesh goiit user id.)

which one is correct???

getting confused..

BEST OF LUCK.

meher sehgal

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3 Jul 2009 13:16:42 IST

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Dear sir,In yr last step there seems a mistake. t should come as (sqrt of v1*V2(/g. This was an assignment quest for me with no answers. I will definitely post the answer as soon I get it. For now the second ans seems more correct. Thanks.

Krishna Gopal Singh

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Joined: 29 Dec 2006

Posts: 4965

4 Jul 2009 00:15:55 IST

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At any time t velocity of first = v1 i - gt j

velocity of second = -v2 i - gt j

Taking dot product and pulling it zero we get t = sqrt(v1*v2)/g

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