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Question

A balloon initially at rest start rising with constant acceleration of 10 m/s2. After 2 s a particle drops from the balloon. After further 2 s, match the following (Take g=10 m/s2 )
Table-1Table-2(A) Height of particle from ground is(P) Zero(B) Speed of particle is(Q) 10 SI units(C) Displacement of particle is(R) 40 SI units(D) Acceleration of particle is (S) 20 SI units

A
(A)(R);(B)(P);(C)(S);(D)(Q)
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B
(A)(R);(B)(Q);(C)(P);(D)(S)
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C
(A)(S);(B)(P);(C)(R);(D)(Q)
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D
(A)(S);(B)(Q);(C)(R);(D)(P)
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Solution

The correct option is A (A)(R);(B)(P);(C)(S);(D)(Q)
Height of the balloon after t=2 s is given by
S1=ut+12(10)22
As the initial velocity of balloon is 0
S1=+20 m
Velocity of balloon after t=2 s is
v=0+20=+20 m/s (Take upward direction as positive)
When the particle is dropped, its initial velocity =+20 m/s
After further 2 s, the displacement of the particle is
S2=20×212×10×22
S2=20 m
Thus height of the particle from ground, S1+S2=40 m
Speed of the particle after further 2 s,v2=2010(2)
v2=0
Since the velocity becomes zero after further 2 s, therefore the particle starts falling freely and hencce the acceleration will act due to graity i.e. 10m/s2.

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