Question

A body falling from height \(H\) takes time \(T\) seconds to reach the ground. The time taken to cover the second half of height is

• A. \(\frac{T}{\sqrt{2}}\)
• B. \(\sqrt{2}T\)
• C. \(\frac{\sqrt{2}-1}{\sqrt{2}}T\)
• D. \(\frac{1}{\sqrt{2}-1}T\)

Hint:

For free fall from rest, distance is proportional to square of time: \(s\propto t^2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A body is released from rest (free fall). We need to analyze the time intervals for covering different segments of the fall.
Under gravity, the displacement is proportional to the square of time (\(h \propto t^2\)).
The time to cover the second half is the total time minus the time taken to cover the first half.
Step 2: Key Formula or Approach:
Equation of motion: \(h = \frac{1}{2}gt^2\) (since \(u = 0\)).
Total time for height \(H\) is \(T\). So, \(H = \frac{1}{2}gT^2\).
Let \(t_1\) be the time to cover the first half height \(H/2\).
Time for second half = \(T - t_1\).
Step 3: Detailed Explanation:

Step A: Relation for total height.
\[ H = \frac{1}{2}gT^2 \implies T = \sqrt{\frac{2H}{g}} \]

Step B: Relation for first half of height.
\[ \frac{H}{2} = \frac{1}{2}g(t_1)^2 \]
\[ \frac{1}{2} \left( \frac{1}{2}gT^2 \right) = \frac{1}{2}g(t_1)^2 \]
\[ t_1^2 = \frac{T^2}{2} \implies t_1 = \frac{T}{\sqrt{2}} \]

Step C: Find time for second half.
\[ t_{\text{second\_half}} = T - t_1 = T - \frac{T}{\sqrt{2}} \]
\[ t_{\text{second\_half}} = T \left( 1 - \frac{1}{\sqrt{2}} \right) \]
\[ t_{\text{second\_half}} = T \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right) \]

Step 4: Final Answer:
The time taken to cover the second half of the height is \(T \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)\).