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

Concept: For a freely falling body starting from rest: \[ s=\frac{1}{2}gt^2 \]

Step 1: Total height is \(H\) and total time is \(T\).
So: \[ H=\frac{1}{2}gT^2 \]

Step 2: Time taken to fall through first half height \(\frac{H}{2}\) is \(t_1\).
So: \[ \frac{H}{2}=\frac{1}{2}gt_1^2 \]

Step 3: Divide this equation by the total height equation. \[ \frac{\frac{H}{2}}{H} = \frac{\frac{1}{2}gt_1^2}{\frac{1}{2}gT^2} \] \[ \frac{1}{2}=\frac{t_1^2}{T^2} \] \[ t_1=\frac{T}{\sqrt{2}} \]

Step 4: Time taken to cover the second half height: \[ t_2=T-t_1 \] \[ t_2=T-\frac{T}{\sqrt{2}} \] \[ t_2=T\left(1-\frac{1}{\sqrt{2}}\right) \] \[ t_2=\frac{\sqrt{2}-1}{\sqrt{2}}T \] Therefore, \[ \boxed{\frac{\sqrt{2}-1}{\sqrt{2}}T} \]