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 t2..
- t1=(√2)t2
- t1=(√2-1)t2
- t2=(√2+1)t1
- t2=(√2-1)t1
The Correct Option is D
Approach Solution - 1
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:
- 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, \(g\).
- 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 \(h_1 = \frac{h}{2}\).
- Plugging \(\frac{h}{2}\) into the equation, we get: \(\frac{h}{2} = \frac{1}{2} g t_1^2\). Simplifying, we find: \(h = g t_1^2\).
- For the second half, time taken is \(t_2\). The ball has traveled a total distance of \(h\) within a total time \(T\), where \(T = t_1 + t_2\).
- At the midpoint (first \(\frac{h}{2}\)), using \(s = \frac{1}{2} g t^2\), we know: \(t_{\text{total}} = \sqrt{\frac{2h}{g}}\).
- 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 \(t_2 = (\sqrt{2}+1)t_1\).
- Thus, \(t_2 = (\sqrt{2} - 1) t_1\).
Therefore, the correct relationship is \(t_2 = (\sqrt{2} - 1) t_1\).
Approach Solution -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
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Concepts Used:
Shortest Distance Between Two Parallel Lines
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)