IIT JEE-AIEEE 2012, Entrance Notifications, Study Tips, Engineering Preparation, Practice Test Papers, Experts Advice

ramyani chakrabarty

Q.19)

mass of the man = M & his initial vel =0

mass of the bag = m

Let the bag be thrown towards left with a vel v so that the vel of man towards right might be V.

Then mv = MV => V = mv / M ------------------(1)

Let the total time taken to reach the ground be t₁ . Now H = building height. So,

H = ( 1/ 2) g t₁ ² => t₁ = ^{[ ]}

( 2H / g ).

Now h = height from the ground at which he throws the bag. Means when he fall a distance ( H - h ) from top , he throws the bag .

Time he will take to fall this distance ( H - h ) is t₂ (say) , where

t₂ = ^{[ ]}

[ 2 ( H - h ) / g ]

now

t₁ - t₂ = ^{[ ]}

( 2H / g ) - ^{[ ]}

[ 2 ( H - h ) / g ]

= ^{[ ]}

( 2 / g ) * [ ^{[ ]}

H - ^{[ ]}

( H -h ) ]

we call this t which is the time in which the man reaches the pond.

now x = horizontal distance covered.

vel of man towards right = V.

so x = V * t

V = x / t and

v = MV / m =( M / m ) * ( x / t )

= ( M / m )* ( x ^{[ ]}

g ) / [ ^{[ ]}

H - ^{[ ]}

( H -h ) ] ^{[ ]}

2

so vel of bag to be imparted is

v = Mx^{[ ]}

g / m [ ^{[ ]}

2H - ^{[ ]}

2( H -h ) ]

As there is no external force in the horizontal dirn, the X co-ordinate of COM will remain in the same posn. So,

0 = [ Mx + m ( -x₁ ) ] / ( M + m )

or, x₁ = Mx / m which is the horizontal dist. covered by bag.

So, the bag will reach the bottom at a distance Mx / m towards left of the line it falls.

cheers !!!

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