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Question Number 15740 by tawa tawa last updated on 13/Jun/17

	Answered by ajfour last updated on 13/Jun/17

	$L e t c l i f f h e i g h t b e h,$ $T o t a l t i m e o f f a l l f r o m c l i f f$ $t o g r o u n d = T;$ $a l s o l e t m e c a l l 1.30 s = t_{0}$ $h = \frac{1}{2} {g t}^{2}$ $\frac{2 h}{3} = \frac{1}{2} g {(t - t_{0})}^{2}$ $u p o n d i v i s i o n,$ $\frac{3}{2} = \frac{t^{2}}{{(t - t_{0})}^{2}} \Rightarrow t - t_{0} = (\sqrt{\frac{2}{3}}) t$ $o r t = \frac{t_{0}}{1 - \sqrt{2 / 3}} = \frac{t_{0} (1 + \sqrt{2 / 3})}{(1 / 3)}$ $h = \frac{1}{2} {g t}^{2} = \frac{g}{2} (9 t_{0}^{2}) (\frac{5}{3} + \frac{2 \sqrt{2}}{\sqrt{3}})$ $= \frac{9.8}{2} (9 \times 1.3 \times 1.3) (\frac{5 + 2 \sqrt{6}}{3})$ $= (4.9 \times 3 \times 1.69) (5 + 2 \times 2.45)$ $= (14.7 \times 1.69) (9.9)$ $h \approx 246 m .$
	Commented by tawa tawa last updated on 13/Jun/17

	$God bless you sir .$
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