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(D) The distance covered by the object in the first $2 \, s$ is given by $h_1 = \frac{1}{2} g t^2 = \frac{1}{2} 	imes 10 	imes 2^2 = 20 \, m$.After

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$(H - 40) \, m$

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(D) The distance covered by the object in the first $2 \, s$ is given by $h_1 = \frac{1}{2} g t^2 = \frac{1}{2} 	imes 10 	imes 2^2 = 20 \, m$.After $2 \, s$,the object is stopped,meaning its velocity becomes $0 \, m/s$ at that instant.It is then released again,so for the next $2 \, s$,it starts from rest with initial velocity $u = 0 \, m/s$.The distance covered in the next $2 \, s$ is $h_2 = \frac{1}{2} g t^2 = \frac{1}{2} 	imes 10 	imes 2^2 = 20 \, m$.The total distance covered from the top is $h_1 + h_2 = 20 + 20 = 40 \, m$.Therefore,the height of the body from the ground is $H - 40 \, m$.

$A$ body is released from the top of a tower of height $H \, m$. After $2 \, s$ it is stopped and then instantaneously released. What will be its height from the ground after the next $2 \, s$?

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