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 determine the relationship between the time taken to fall the first half of the distance and the second half of the distance for a freely falling ball (under gravity). Let's break down the solution step-by-step: 

1. We are given that a ball is released from a height \(h\). When the ball is dropped from a height, it accelerates downwards under the influence of gravity, \(g\).
2. The time taken to cover the first half of the distance \(\frac{h}{2}\) is \(t_1\). Using the kinematic equation for uniformly accelerated motion: \(h_1 = \frac{1}{2} g t_1^2\), where \(h_1 = \frac{h}{2}\).
3. Plugging \(\frac{h}{2}\) into the equation, we get: \(\frac{h}{2} = \frac{1}{2} g t_1^2\). Simplifying, we find: \(h = g t_1^2\).
4. For the second half, time taken is \(t_2\). The ball has traveled a total distance of \(h\) within a total time \(T\), where \(T = t_1 + t_2\).
5. At the midpoint (first \(\frac{h}{2}\)), using \(s = \frac{1}{2} g t^2\), we know: \(t_{\text{total}} = \sqrt{\frac{2h}{g}}\).
6. Using the known fact for falling objects, \(t_1 = \frac{\sqrt{2}-1}{\sqrt{2}+1} \cdot t_2\), approximately: \(t_1 = \frac{t_{\text{total}}}{\sqrt{2}+1}\), which becomes \(t_2 = (\sqrt{2}+1)t_1\).
7. Thus, \(t_2 = (\sqrt{2} - 1) t_1\).

Therefore, the correct relationship is \(t_2 = (\sqrt{2} - 1) t_1\).

Answer 2

\(t_1=\frac{\sqrt2⋅\frac{H}{2}}{g}=\frac{\sqrt{H}}{g}\)
\(t_2=\frac{\sqrt{2H}}{g}−t_1\)
\(⇒t_2=\frac{\sqrt{2H}}{g}−\frac{\sqrt{H}}{g}\)
\(⇒t2=√Hg{√2−1}\)
\(⇒t2=(√2−1)t1\)
So, the correct option is (D): t2 =(√2−1)t1