A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is

The correct option is 24.5 m.

Let us analyze the motion of both bodies step-by-step under the influence of gravity. Let the acceleration due to gravity be $g=9.8{\text{m/s}}^2$ acting downwards.

The distance $s$ traveled by a body falling freely from rest in time $t$ is given by the second equation of motion:
$s=\frac{1}{2}g{t}^2$

Let the first body be released at time $t=0$ seconds.
The second body is released exactly one second later, i.e., at $t=1$ second.

We are asked to find the separation between the two bodies two seconds after the release of the second body.
At this instant:
- The time elapsed for the second body since its release is $t_2=2\text{seconds}$.
- The time elapsed for the first body since its release is $t_1=2+1=3\text{seconds}$.

Now, let us calculate the distance fallen by each body from the initial height:

For the first body:
$s_1=\frac{1}{2}g{t}_1^2=\frac{1}{2}\times 9.8\times 3^2=4.9\times 9=44.1\text{m}$

For the second body:
$s_2=\frac{1}{2}g{t}_2^2=\frac{1}{2}\times 9.8\times 2^2=4.9\times 4=19.6\text{m}$

The separation between the two bodies is the difference in the distances they have fallen:
$\text{Separation}=s_1-s_2=44.1\text{m}-19.6\text{m}=24.5\text{m}$

Therefore, the separation between the two bodies two seconds after the release of the second body is 24.5 m.