A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then be
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\(h - ut + \frac{1}{2}gt^{2} = 1 - 0/t + \frac{1}{2}gt^{2} = t - \sqrt{2/g}\)
Velocity after travelling 1 m distance
\(v^{2} = u^{2} + 2gh \Rightarrow v^{2} = (0)^{2} + 2g \times 1 = v = \sqrt{2g}\)
For second 1 meter distance \(1 - \sqrt{2g} \times t_2 + \frac{1}{2} g t_2^2 = g t_1^2 + 2 \sqrt{2g} t_2 - 2 = 0\)
\(\epsilon_2 = \frac{2 \sqrt{2} g - y \sqrt{8} g + 8 g}{2 g} \cdot \frac{\sqrt{2} \pm 2}{\sqrt{g}}\)
Taking +ve sign \(t_{2} = \frac{2 - \sqrt{2}}{\sqrt{g}}\)
\(\cdot\) \(\frac{t_1}{t_2} = \frac{\sqrt{2} \sqrt{7} g}{(2 - \sqrt{2}) \sqrt{g}} = \frac{1}{\sqrt{2} - 1}\) and so on.
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