Step 1: Understanding the Concept:
For a body falling from rest, the displacement $s$ is proportional to the square of time $t$ according to the equation $s = \frac{1}{2}gt^2$. We need to find the total time $T$ for height $H$ and the time $t_1$ for the first half height $H/2$. The time for the second half is the difference between these two.
Step 2: Key Formula or Approach:
1. Total time: $H = \frac{1}{2}gT^2 \implies T = \sqrt{\frac{2H}{g}}$.
2. Time for first half: $\frac{H}{2} = \frac{1}{2}gt_1^2 \implies t_1 = \sqrt{\frac{H}{g}}$.
Step 3: Detailed Explanation:
From the equations above, we can see that:
\[ t_1 = \frac{1}{\sqrt{2}} \sqrt{\frac{2H}{g}} = \frac{T}{\sqrt{2}} \]
The time taken for the second half of the height ($\Delta t$) is:
\[ \Delta t = T - t_1 = T - \frac{T}{\sqrt{2}} \]
Factor out $T$:
\[ \Delta t = T \left( 1 - \frac{1}{\sqrt{2}} \right) = T \left( \frac{\sqrt{2}-1}{\sqrt{2}} \right) \]
Step 4: Final Answer:
The time taken for the second half is \( \left( \frac{\sqrt{2}-1}{\sqrt{2}} \right) T \).