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A balloon rise up (initial velocity is zero) with constant net acceleration of $10 m / s^{2}$ . 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 the ground p) Zero
b)Speed of the body Q)10SIunits
c)Displacement of the body in 2s after it drops from the balloon R)40SI units
d) Acceleration of the body S)20 SI units

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$A - R, B - P, C - S, D - Q$

$A - P, B - R, C - S, D - Q$

$A - Q, B - R, C - S, D - P$

$A - S, B - P, C - R, D - Q$

answer is A.

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Detailed Solution

After $2 \sec, h = \frac{1}{2} a t^{2}$
$h = \frac{1}{2} \times 10 \times 4 = 20$
Velocity of balloon and stone after $2 \sec$ is $u = a t \Rightarrow u = 10 \times 2 = 20 m / s$
Velocity of stone after further after $v = u - g t = 20 - 20 = 0$ displacement after further $2 \sec$
$S = u t - \frac{1}{2} g t^{2} = 20 m$ from ground its Position is $20 + 20 = 40 m$

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