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A balloon rises up (initial velocity is zero) with constant net acceleration $10 m / s^{2} .$ After 2 seconds, a body drops from the balloon. After further 2 seconds, match the column-I with column-II ( $g = 10 m / s^{2}$ ).
Column-I Column-II
A) Height of body from ground P) zero
B) Speed of the body Q) 10 S.I. units
C) Displacement of the body in 2s after it drops from the balloon R) 40 S.I. units
D) Acceleration the body S) 20 S.I. units

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A – R, B – S, C – R, D – Q

A – S, B – Q, C – P, D – R

A – P, B – R, C – Q, D – P

A – R, B – P, C – S, D – Q

answer is D.

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Detailed Solution

After 2 seconds:
Velocity of the balloon and stone after 2 secs is $u = a t = 10 \times 2 = 20 m / s$
Velocity of the stone $v = u - g t = 20 - 10 \times 2 = 0 m / s$
Displacement of stone after further 2 secs is $S = u t - \frac{1}{2} a t^{2} = (20 \times 2) - \frac{1}{2} \times 10 \times 2^{2} = 20 m$
After 2 sec, $h = \frac{1}{2} a t^{2} = \frac{1}{2} \times 10 \times 2^{2} = 20 m$
From ground its point= 20+20 = 40m

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