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A body dropped from the top of tower falls through 200 m during last 5 seconds of its fall. Find height of the tower. |
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Answer» In case of free fall displacement of the motion \(S = \frac 12 gt^2\) nth four second (n - 5)th = tn-5 given (tn - tn-5) = 5 Displacement in last five second \(S_{n , n-5} = \frac 12 g (t_n - t_{n -5})^2\) \(200 = \frac 12 \times 10\times (t_n - t_{n - 5})^2\) \(40 = (t_n - t_{n - 5}) (t_n + t_{n - 5})\) \(40 = 5 (t_n + t_{n - 5})\) \(8 = t_n + t_{n - 5}\) Solving equation (1) and (2) \(\begin{array}{r}t_n + t_{n-5} = 8\\t_n - t_{n - 5} = 5\\\hline 2t_n= 13\end{array}\) \(t_n = \frac{13}2\) \(t_n = 6.5 \) sec \(t_{n-5} = 5 + 6.5\) \(t_{n-5} = 11.5\) Height of the building \(S = \frac 12 gt^2\) \(S = \frac 12 \times 10\times (6.5)^2\) \(S = 5\times 42.25\) \(S = 211.25\) |
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