Question

A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t_2..

• A. t₁=(√2)t₂
• B. t₁=(√2-1)t₂
• C. t₂=(√2+1)t₁
• D. t₂=(√2-1)t₁

Answer 1

The Correct Option is D

   To solve this problem, we need to calculate the time taken for a ball to complete two different halves of its fall from a height \(h\). Let's break down the steps to find the relation between the time for the first half, \(t_1\), and the second half, \(t_2\).

1. The ball is released from rest, hence its initial velocity \(u = 0\).
2. The formula for the distance \(s\) covered under uniform acceleration \(a\) is given by:

\(s = ut + \frac{1}{2}at^2\)

1. For a freely falling object, the acceleration \(a\) is \(g\) (acceleration due to gravity).
2. The total height is \(h\). The time to fall through the entire height is given by:

\(h = 0 \times T + \frac{1}{2}gT^2\), which simplifies to \(T = \sqrt{\frac{2h}{g}}\)

1. The first half of the distance is \(\frac{h}{2}\). Using \(s = \frac{1}{2}gt_1^2\), we have:

\(\frac{h}{2} = \frac{1}{2}gt_1^2 \Rightarrow h = gt_1^2 \Rightarrow t_1 = \sqrt{\frac{h}{g}}\)

1. For the second half of the fall, the distance is also \(\frac{h}{2}\), and the time for the entire fall is \(T\), so for the second half:

\(t_2 = T - t_1 = \sqrt{\frac{2h}{g}} - \sqrt{\frac{h}{g}}\)

This simplifies to \(t_2 = (\sqrt{2} - 1)\sqrt{\frac{h}{g}} = (\sqrt{2} - 1)t_1\)

Thus, the correct relation between \(t_1\) and \(t_2\) is: \(t_2 = (\sqrt{2} - 1)t_1\).

This confirms the provided correct answer option:

t₂=(√2-1)t₁